1> 



LIBRARY OF CONGRESS. 5 



t UNITED STATES OP AMERICA, t 






\ 



A MANUAL 



ELEMENTARY 

GEOMETRICAL DRAWING, 

INVOLVING THREE DIMENSIONS. 



DESIGNED FOR USE IN HIGH SCHOOLS, ACADEMIES, ENGINEERING SCHOOLS, ETC. J 
AND FOR THE SELF-INSTRUCTION OF INVENTORS, ARTIZANS, ETO. 



In irioe HDitrisions. 

DIV. I. ELEMENTARY PROJECTIONS. 

DIV. II. DETAILS OF CONSTRUCTIONS IN MASONRY, WOOD, AND METAL. 

DIV. III. RUDIMENTARY EXERCISES IN SHADES AND SHADOWS. 

DIV. IV. ISOMETRICAL DRAWING. 

DIV. V. ELEMENTARY STRUCTURAL DRAWING. 



By S. EDWARD WARREN, C. E. 

OF DESCRIPTIVE GEOMETRY, ETC., IN THE RENSSELAER POLYTECHNIC INSTITUTE, AND AUTHOB 

OF "ELEMENTARY PLANE PROBLEMS;" 1 "DRAFTING INSTRUMENTS, ETC.;" 

"ELEMENTARY LINEAR PERSPECTIVE;" "DESCRIPTIVE 

GEOMETRY," AND "SHADES AND 

SHADOWS." 



Third. Edition, .Revised and Enlarged. 

f NEW YORK: 
JOHN WILEY & SON, 535 BROADWAY. 

1867. 



Entered according to Act of Congress, in the year eighteen hundred and sixty-one, by 

S. EDWAED WAEEEU, 

In the Clerk's office of the District Court of the United States for the Northern District of 

New York. 



K 



The New York Printing Company, 

81, 83, and 85 Centre Street^ 

New York. 



V 



*v 



CONTENTS. 



PA SI 

Note to the Third Edition vi 

Preface , vii 

Preliminary Notes x 

DIVISION FIRST. 

PROJECTIONS. 

Chapter I. — First Principles. 

§ I, — The purely geometrical or rational theory of projections 1 

§ II. — Of the relations of lines to their projections 8 

§ III. — Physical theory of projections 5 

§ IV. — Conventional mode of representing the two planes of projection, 
and the two projections of any object upon one plane ; viz. 

the plane of the paper. 5 

§ V. — Of the conventional direction of the light ; and of the position 

and use of heavy lines 6 

§ VL— Notation 7 

§ VII. — Of the use of the method of projections 8 

Chapter II. — Projection of lines. Problems in right projection ; and in oblique 
projection, snowing two sides of a solid right angle. (Thirty- 
two problems.) 

§ I. — Projections of straight lines 9 

§ II. — Right projections of solids 12 

§ III. — Oblique projections showing two sides of a solid right angle. . . 14 
§ IV. — Special Elementary Intersections and Developments 23 



DIVISION SECOND. 

DETAILS OF CONSTRUCTIONS IN MASONRY, WOOD, AND METAL. 

Chapter I. — Constructions in Masonry. 

§ L — General definitions and principles applicable both to brick and 

stone work , . , 32 



IV CONTENTS. 

PAGB 

§ II. — Brick work 32 

§ III— Stone work 35 

Chapter II. — Constructions in Wood. 

§ I. — General remarks. (Explanation of Scales.) 3*7 

§ II.— Pairs of timbers whose axes make angles of 0° with each other. 40 
§ III. — Combinations of timbers whose axes make angles of 90° with 

each other 43 

§ IV. — Miscellaneous combinations. (Dowelling, &c.) 48 

§ Y. — Pairs of timbers which are framed together obliquely to each 

other ; 4*7 

§ VI. — Combinations of timbers whose axes make angles of 180° with 

each other 43 

Chapter III. — Constructions in Metal 53 

Cage valve of a locomotive pump. Metallic packing for stuff- 
ing boxes, &c. 54 



DIVISION THIRD. 

RUDIMENTS OF SHADES AND SHADOWS. 



Chapter I. — Shadows. 

§ I. — Observed facts and practical rules 59 

§ II.— Problems. (Twelve.) 63 

Chapter II. — Shading 10 

(Hexagonal prism ; cylinder; cone, and sphere.) 



DIVISION FOURTH. 

ISOMETRICAL drawing and cabinet PROJECTIONS. 

Chapter I. — First Principles of Isometrical Drawing 71 

" II. — Problems involving only isometric lines 81 

" III. — Problems involving non-isometrical lines 84 

" IV. — Problems involving the construction and equal division of circles in 

Isometrical Drawing 88 

(Approximate construction of the isometrical drawing of a 
circle.) (Nineteen problems.) 
" V. — Cabinet Projections 95 



CONTENTS. V 

PAGB 

DIVISION" FIFTH. 

ELEMENTARY STRUCTURAL DRAWING. 

Chapter I. — Stone structures 101 

A brick segmental arch 101 

A semi-cylindrical culvert, with vertical cylindrical wing- 
walls 103 

Chapter II. — Wooden structures 107 

A king post truss 107 

A queen post truss-bridge 108 

An architectural model, with shadows » , . 110 

Chapter III. — Iron constructions. 114 

A railway track — frog, chair, &c 114 

A hydraulic ram 116 



NOTE TO THE THIRD EDITION. 



At the suggestion of several friends, I have at last been induced 
to add a fourth section to Division I, Chapter II., embracing 
some useful special intersections, &c, and one additional plate. 
I have waited, partly, till a vision of the seemingly best selec- 
tion of figures as to novelty, variety, and utility, for a single 
plate only, should present itself; and partly, doubting whether 
all such problems should not be referred only to the introduction 
to a contemplated work on Machine Drawing. But a few, of 
the larger number necessary there, may be found useful here, 
especially to sheet metal workers, and young machinists, and are 
accordingly inserted. I am not aware that the pleasing and 
useful constructions-Figs. 8 and 9, PI. III. A,-can anywhere 
else be found in a work so readily obtainable as the present. 

An important addition has also been made to Division IV., in 
the chapter on " Cabinet Projections;" which are often preferable 
to Isometrical Drawings. And special pains have been taken 
to improve the treatment of the subject of Shadows. 

Besides these principal changes, numerous minor corrections 
and improved paragraphs, have been scattered through the pre- 
sent edition : the result of a careful revision, designed to make 
my work, as a whole, more acceptable to all having occasion to 
use it. 

R. P. I, Nov., 1866. 



PREFACE. 



Experience in teaching shows, that correct conceptions of the 
forms of objects having three dimensions, are obtained with 
considerable difficulty by the beginner, from drawings having 
but two dimensions, especially when those drawings are neither 
"natural"— that is " pictorial "—nor shaded, so as to suggest 
their form ; but are artificial, or " conventional," and are merely 
" skeleton," or unshaded, line drawings. Hence moderate experi- 
ence suggests, and continued experience confirms, the propriety 
of interposing, between the easily understood drawings of pro- 
blems involving two dimensions, and the general course of pro- 
blems of three dimensions, a rudimentary course upon the methods 
of representing objects having three dimensions. This much 
may be said in defence of Division I. of the present volume. 

Experience again proves, in respect to the drawing of any 
engineering structures that are worth drawing, that it is a great 
advantage to the draftsman to have— 1st, some knowledge of the 
thing to he drawn, aside from his knowledge of the methods of 
drawing it ; and 2d, practice in the leisurely study of the graphical 
construction of single members or elements of a piece of framing, 

or other structure. 

The truth of the second of the preceding remarks, is further 
apparent, from the fact that in entering at once upon the draw- 
ing of whole structures, three evils ensue, viz.— 1st, Confusion 
of ideas, arising from the mass of new objects (the many different 
parts of a structure) thrown upon the mind at once ; 2d, Loss 
of time, owing to repetition of the same detail many time* in 



yiii PREFACE. 



the same structure ; and 3d, Waste of drawings, as well as of 
time, through poor execution, which is due to insufficient previ- 
ous practice. 

Hence Division II. is rilled up with a tolerably liberal and 
comprehensive collection of details of structures, each one of 
which affords a useful problem, illustrative of some one feature 
of constructive art and of graphical procedure. 

Here it may be said, that the department of instruction, part 
of which is embodied in this work, has involved in it a sub- 
department, viz., that of Descriptive Constructions, i.e. a popular 
descriptive study of the ways in which many common things are 
made and put together, or in which they operate. 

The classification of Chap. II., Div. II., is believed to be a 
happy one, as readily suggesting various forms of framings. 

Since my object is not so much to acquaint the student solely 
with the works of the authors or artizans of other nations, as to 
give them just those specimens of American methods of con- 
struction which are not usually found in books, I must refer 
them to foreign works for examples of foreign modes of con- 
struction — not that such examples should be disparaged, but 
because it is intended that this work shall add something to our 
stock of book knowledge of the subjects of which it treats, 
rather than merely repeat what has been before published. 

Division III. contains a short special study of Shadows, taken 
up in the form of the study of a few individual problems of a 
practical character, assuming that the light comes only in the 
simple conventional direction, commonly agreed upon by drafts- 
men, and saying nothing of the possibility of allowing the light 
to be taken in any other direction. The problems of this Divi- 
sion are intended to be sufficient, when fully understood, to 
enable the student to construct accurately, or by a sufficiently 
near approximation, all the shadows he will meet with in 
Geometrical Drawing, till he shall pursue a general course on 
shades and shadows. 

Division IV. contains a sufficient exposition of the few and 



PREFACE. IX 

simple principles of Isometrical projection and drawing, to meet 
all the ordinary wants of the student or draftsman. 

Division Y. includes examples of a few simple structures, to 
fulfil the threefold purpose of affording occasion for learning the 
names of parts of structures ; for practice in the combination of 
details into whole structures ; and for profitable review practice 
in execution. 

Finally, since experience has dictated so much of this preface, 
it shall dictate also these closing remarks. First : Classes will 
generally be found to take a lively interest in the subjects of 
this volume — because of their freshness to most learners, as new 
subjects of interesting study — because of the variety and brevity 
of the topics, which causes frequent change of the subject of 
study (such change being thought quite desirable by most 
learners) — and because of the compactness, neatness, and beaut} 
of the volume which is formed by binding together all the plates 
of the course, when they are well executed. Second : As to 
the use of this volume, it is intended that there should be formal 
recitations of the problems in the 1st, 3d, and 4th divisions, with 
graphical constructions of a selection of the same or similar 
ones ; and occasional interrogations mingled with the graphical 
constructions of the practical problems of the remaining divi- 
sions. Remembering that excellence in mere execution, though 
highly desirable and to be encouraged, is not, at this stage of 
the student's progress, the sole end to be attained, a small num- 
ber of graphic exercises will be sufficient. The student may, in 
place of a tedious course of finished drawings, be called on fre- 
quently to describe, by the aid of pencil or blackboard sketches, 
how he would construct drawings of certain objects — either those 
given in the several Divisions of this volume, or other similar 
ones proposed by his teacher. 

In a course of school education, the full delivery to the stu- 
dent of the greatest number of distinct ideas, with the least amount 
of merely manual lahor^ should be the teacher's general aim. 



PRELIMINARY NOTES. 



As beginners not seldom find peculiar difficulties at the outset 
of the study of projections, the removal of which, however, 
makes subsequent progress easy, the following special explana- 
tions are here prefixed. 

I. Figs. 1, 2, 3, 5, and 15, of PL I. are pictorial diagrams, 
used for illustration in place of actual models. Thus, in Fig. 3, for 
example, MHi represents a horizontal square cornered plane sur- 
face, as a floor. MGY represents a vertical square cornered 
plane surface, as a wall, which is therefore perpendicular to 
MHi. P represents any point in the angular space included by 
these two planes. ~Pp represents a line from P, perpendicular to 
the plane MHi, and meeting it at^>. Pp' represents a line from 
P, perpendicular to the plane MGY, and meeting it at p' . Then 
Pp and Pp' are called the projecting lines of P. The point p is 
called the horizontal projection of P, and p' the vertical projection 
of P. The projecting lines of any point or of any body, as in 
Fig. 1, are perpendicular to the planes, as MH/ and MGY, 
which are called planes of projection. 

II. In preparing a lesson from this work, the object of the 
student is, by no means, to commit to memory the figures, but 
to learn, from the first principles, and -subsequent explanations, 
to see in these figures the realities in space which they represent; so 
as to be able, on hearing the enunciation of any of the problems, 
to solve it from a clear understanding of the subject, and not 
"by rote" from mere memory of the diagrams. The student 
will be greatly aided in so preparing his lessons, by working 
out the problems, in space, on actual planes at right angles to 
each other, as on the leaves of a folding slate, when one slate is 
placed horizontally and the other vertically. 



A MANUAL 



ELEMENTARY GEOMETRICAL DRAWING. 



DIVISION FIRST, 

PROJECTIONS. 



CHAPTER I. 

FIRST PRINCIPLES. 



§ I. The purely Geometrical or Rational Theory of Projections, 

1. It is always easy to make a graphic solution of any problem 
involving but two dimensions, or a graphical construction of any 
object having but two dimensions, upon a plane surface, as a sheet 
of paper ; since, upon a surface having two dimensions, any figure 
having but two dimensions can have both of those dimensions repre- 
sented in their true size and relative position. 

2. The preceding article leads to the conclusion, that upon a 
plane surface, which has but two dimensions, no more than two 
dimensions can be represented in their true size and relative posi- 
tion. But when we consider the vast number of geometrical mag- 
nitudes, as cylinders, cones, spheres, &c, which have three dimen- 
sions, it appears probable, and actually is true, that the study of 
those magnitudes gives rise to a great many problems involving 
three dimensions. It appears, then, that there are problems of 
three as well as of two dimensions, and also that more than one 
plane surface is necessary to enable us to represent, both com- 
pletely and accurately, bodies having three dimensions. 

3. What then is the number and the relative position of the planes 
which will enable us to represent all the dimensions of any geome- 
trical solid, in their real size, on those planes ? To assist in answer- 
ing this question, reference may be made to PI. I. Fig. 1. Let 
ABCFED be a regular square-cornered block, whose length is AB ; 
breadth, AD ; and thickness, AC ; and let MN be any horizontal 



2 FIRST PRINCIPLES. 

plane below it and parallel to its top surface ABED. If now the 
four vertical edges of the block, at A, B, E, and D, be produced to 
meet this plane, they will meet it in points — as a, b, e, and d. By 
joining these points, it is evident that a figure — abed — will be 
formed, which will be equal to the top surface of the block, and will 
be a correct representation of the length and breadth of that toj 
surface — i. e. of the length and breadth of the block. Similarly,, 
let MP be a vertical plane, behind the block and parallel to its front 
face, ABCF. Produce the four horizontal edges, through A, B, C, 
and F, till they meet the vertical plane, MP, and connect the points, 
a\ b\ c\ and f\ thus found. The figure, a'c'b'f, so found, will 
evidently be equal to the front face ABCF of the block, and will 
therefore correctly represent the length and thickness of the block. 

4. From the foregoing article, the following fundamental princi- 
ples are deduced. First: Two planes, at right angles to each 
other, are necessary to enable us to represent, fully, the three dimen- 
sions of a solid. Second: In order that those dimensions shall be 
seen in their true size and relative position, they must be parallel to 
that plane on which they are shown. Third: Each plane shows 
two of the dimensions of the solid; viz., the two which are parallel 
to it, and that dimension which is thus shown twice, is the one 
which is parallel to both of the planes. Thus AB, the length, and 
AD, the breadth, are shown on the plane MN ; and AB, the length 
again, and AC, the thickness, are shown on the plane MP. 

5. From Art. (3), the following definitions naturally arise. If a 
body be thrown directly forward, as across a table, all its points will 
describe parallel straight lines, while the body will be said to be pro- 
jected. Hence, as the lines of each group which are drawn through 
the edges of the block, PI. I., Fig. 1, to produce the figures abed 
and a'b'c'f, are parallel, the block is naturally said to be projected 
upon the planes MN and MP; and the figures abed and a'b'c'f 
are called the projections of the block. The planes, containing 
these figures, are hence naturally called planes of projection — MN, 
being horizontal, is called the horizontal plane of projection, and 
MP, being vertical, is called the vertical plane of projection. 

The projections take their particular names from the planes in 
which they are found. Thus, abed is the horizontal projection of the 
block, and a'b'c'f is its vertical projection. The same idea is 
expressed otherwise, by saying, that the block is horizontally 
projected in the figure abed, and that it is vertically projected in the 
figure a'b'c'f '. 

The system of lines, of which Aa is one, may be called horizon- 



FIRST PRINCIPLES. 3 

tally projecting lines, and the group of which Aa r is one, may be 
called vertically projecting lines. These projecting lines are always 
perpendicular to the planes of projection. Finally, MR, the inter- 
section of these planes, is called the ground line. 

6. The preceding definitions enable us to express a fourtlriuii- 
damental principle, more briefly than could otherwise be done; 
viz. The height of the vertical projection of a point above the 
ground line, is equal to the height of the point itself, in space, 
above the horizontal plane : and the perpendicular distance of the 
horizontal projection of a point from the ground line, is equal to 
the p>erpendicular distance of the point itself, in front of the vertical 
'plane. Thus; PI. L, Fig. 1, ad'—Ka' and a'a"—Aa. 

7. These four fundamental principles, with the accompanying 
definitions, are the foundation of the subject of projections, but, by 
attending carefully to PI. I., Fig. 2, some additional special prin- 
ciples may be discovered, which are frequently applied in practice. 
PL I., Fig. 2, is a pictorial model of a pyramid, Vcdeg, and of its 
two projections. The face, Vcd, of the pyramid, is parallel to 
the vertical plane, and the triangle, Xa5, is equal and parallel to 
Ycd, and a little in front and at one side of it. By first conceiv- 
ing, now, of the actual models, which are, perhaps, represented as 
clearly as they can be by mere diagrams, in PI. I., Figs. 1 and 2 ; 
and then by attentive study of those figures, the facts stated in the 
next two articles will be fully comprehended. 

§ II. — Of the Relations of Lines to their Projections. 

8. Relations of single lines to their projections. 

a. A vertical line, as AC, PI. I., Fig. 1, has, for its horizontal 
projection, a point, a, and for its vertical projection, a line a'c', 
perpendicular to the ground line, and equal and parallel to the line 
AC, in space. 

b. A horizontal line, as AD, which is perpendicular to the verti- 
cal plane, has, for its horizontal projection, a line, ad, perpen- 
dicular to the ground line, and equal and parallel to the line, AD, 
in space ; and for its vertical projection a point, a'. 

e. A horizontal line, as AB, which is parallel to both planes of 
projection, has, for both of its projections, lines ab and a'b', which 
are parallel to the ground line, and equal and parallel to the line, 
AB, in space. 

d. A horizontal line, as BD, which makes an acute angle with 
the vertical plane, has, for its horizontal projection, a line, bd, 
which makes the same angle with the ground line that the line, BD, 



FIEST PRINCIPLES. 



makes with the vertical plane, and is equal and parallel to the line 
itself (BD) ; and has for its vertical projection a line b'a', which is 
parallel to the ground line, but shorter than BD, the line in space. 

e. An oblique line, as BC, PL I., Fig. 1, or Yd, PI. L, Fig. 2, 
which \s parallel to the vertical plane, has, for its vertical projection, 
a line b'c', or v'd', which is equal and parallel to itself, and for its 
horizontal projection, a line ba or vd, parallel to the ground line, 
but shorter than the line in space. 

/. An oblique line, as Yg, PI. I., Fig. 2, which is oblique to both 
planes ofprojectio?i, has both of its projections, v'd' and vg, oblique 
to the ground line, and shorter than the line itself. 

g. An oblique line, as AH, PI. I., Fig. 1, which is oblique to both 
planes of projection, but is in a plane ACDH, perpendicular to 
both of those planes, has both of its projections, a'c' and ad, perpen- 
dicular to the ground line, and shorter than the line itself. 

h. A line, lying in either plane of projection, coincides with its 
projection on that plane, and has its other projection in the ground 
line. See cd — c'd', the projections of cd, PI. I., Fig. 2. 

9. Remark. A general principle, which it is important to be 
perfectly familiar with, is embodied in several of the preceding 
examples; viz. When any line is parallel to either plane of projec- 
tion, its projection on that plane is equal and parallel to itself, and 
its projection on the other plane is parallel to the ground line. 

10. The preceding remark serves to show how to find the true 
length of a line, when its projections are given. When the line, as 
Yg, PI. I., Fig. 2, is oblique to both planes of projection, its length, 
Yg, is evidently equal to the hypothenuse of a right-angled triangle, 
of which the base is vg, the horizontal projection of the line, and the 
altitude is Yv, the height of the upper extremity, V, above the 
horizontal plane. When the line, as AH, PI. I., Fig. 1, does not 
touch either plane of projection, it is evidently equal to the hypo- 
thenuse of a right-angled triangle, of which the base, CH, equals 
the horizontal projection, ad, and the altitude, AC, equals the 
difference of the perpendiculars, Aa and He?, to the horizontal plane. 

In the same way, it is also true that the line, as Yg, PI. I., Fig. 2, 
is the hypothenuse of another right-angled triangle, whose base 
equals the vertical projection, v'd' , and whose altitude equals the 
difference of the perpendiculars, Yv' and d'g, from the extremities 
of the line to the vertical plane of projection. 

1 1 . Relations of pairs of lines to their projections. These rela- 
tions, after the full notice now given of the various positions of 
single lines, may be briefly expressed as follows. 



FIRST PRINCIPLES. 5 

a. A pair of lines which are equal and parallel in space, and also 
parallel to a plane of projection, as AB and CF, PI. I., Fig. 1, or Vc 
and Xa, PI. L, Fig. 2, have their projections on that plane — a'b' 
and cf, PI. L, Fig. 1, or v'c' and x'a', PI. I., Fig. 2 — equal and 
parallel — to each other, and to the lines in space. 

b. A pair of lines which are equal and parallel in space, but not 
parallel to a plane of projection, will have their projections on that 
plane equal and parallel to each other, but not to the lines in 
space. 

c. Parallel lines make equal angles with either plane of projec- 
tion ; hence it is easy to see that lines not parallel to each other — 
as Yd and Vc, or Yg and Ye, PI. I., Fig. 2 — but which make 
equal angles with the planes of projection, will have equal projec- 
tions on both planes — i.e. v'd'—v'c' and vg~ve, also vd—vc. 

§ III. — Physical Theory of Projections. 

12. The preceding articles comprise the substance of the purely 
geometrical or rational theory of projections, which, strictly, is 
sufficient ; but it is natural to take account of the physical fact that 
the magnitudes in space and their representations, both address 
themselves to the eye, and to inquire from %ohat distance and in 
what direction the magnitudes in PL I., Figs. 1 and 2, must be 
viewed, in order that they shall appear just as their projections 
represent them. Since the vertically-projecting perpendiculars, Q, 
regarded as rays of light, reflected from the block, Fig. 1, could 
only meet in the eye at an infinite distance in front of the vertical 
plane, we conclude that the vertical projection of an object repre- 
sents that object, in respect to form, as it would appear to the eye 
situated at an infinite distance from it, and looking in a direction 
perpendicular to the vertical plane of projection. Reasoning simi- 
larly in reference to the group, S, we conclude that the horizontal 
projection of an object presents the same appearance, as respects 
form, that would be afforded by the object itself, when viewed by 
the eye at an infinite distance above it, looking in a direction per- 
pendicular to the horizontal plane. 

§ IV. — Conventional Mode of representing the two Planes of Pro- 
' jection, and the two Projections of any Object upon one plane 
— viz. the Plane of the Paper. 

13. In practice, a single flat sheet of paper represents the two 
planes of projection, and in the following manner. The vertical 
plane, MY, PI. I., Fig. 3, is supposed to revolve backwards, as 



6 FIRST PRINCIPLES. 

shown by the arcs ru and Vt, till it coincides with the horizontal 
plane produced at M ut G. Hence, drawing a line from right to 
left across the paper, to represent the ground line, MG, all that 
part of the paper above or beyond such a line will represent the 
vertical plane of projection, and the part below it the horizontal 
plane of projection. 

14. Elementary Rational Geometry teaches that two lines, as 
PP', Pp, PI. I., Fig. 3, which intersect, fix the position of the plane 
containing them. It also teaches that if from a point in space, as 
P, two lines P — P', P— p, be drawn — one and the other perpendi- 
cular to one and the other of two planes, MV and MH, which are 
at right angles to each other — then the plane, YP' p"p, of these 
lines, will be perpendicular to the two given planes, and will inter- 
sect those planes, in lines Y'—p" and p"—p, which will be perpen- 
dicular to the intersection, MG, of the planes, at the same point, p". 

This conclusion being reached, it can now be expressed more 
briefly by saying, that the plane of the projecting lines of any point, 
intersects the planes of projection, in lines which are perpendi- 
cular to the ground line at the same point. 

15. Now, when the vertical plane revolves about the line MG, as an 
axis, the point p\ being in the axis, must remain fixed, and any other 
point, as P', must revolve in a plane perpendicular to MG. Hence 
the line P' — p" will revolve — with the plane MV — to the position 
p' — -p\ still remaining perpendicular to the ground line. Hence 
we have this fundamental principle of graphic constructions, viz. 
Tfie two projections p, p\ PI. I., Figs. 3 or 4, of a point in space, 
are always in a line, pp\ perpendicular to the ground line. 

§ V . — Of the Conventional Direction of the Light / and of the 
Position and Use of Heavy Lines. 

16. A single topic from Division III. must here be anticipated, 
viz., that which refers to the assumption of a certain direction for 
the light. Without going into the subject fully, it will be sufficient 
to say, that as one stands facing the vertical plane of projection, 
the light is assumed to come from behind, and over the left 
shoulder, in such a direction that each of its projections makes an 
angle of 45° with the ground line, as shown in PI. I., Fig. 6. 

17. The practical effect of the preceding assumption in reference 
to the light, is, that upon a body of the form and position shown in 
PI. I., Fig. 5, for example, the top, front, and left hand surfaces— 
i.e. the three seen in the Fig. — are illuminated, while the other 
three faces of the body are in the shade. 



FIRST PRINCIPLES. 7 

18. The practical rule by which the direction of the light and 
its effect are indicated in the projections, is, that all those visible 
edges of the body in space, which divide the light from the dark 
surfaces, are made heavy in projection. 

19. To illustrate : The edges BC and CD of the body in space, 
Pi. I., Fig. 5, divide light from dark surfaces, and are seen in 
looking towards the vertical plane, and hence are made heavy in 
vertical projection, as seen at b'c' and c'd' . BK and KF divide 
illuminated from dark surfaces, and are seen in looking towards 
the horizontal plane, and are therefore made heavy in the hori- 
zontal projection, as shown at bh and Jcf. 

20. By inspection, it will be seen that the following simple rule 
in reference to the position of the heavy lines on the drawings, may 
be deduced, as an aid to the memory. In all ordinary four-sided 
prismatic bodies, placed with their edges respectively parallel and 
perpendicular to the planes of projection, or nearly so, the right 
hand lines, and those nearest the ground line, of both projections, 
are made heavy. 

21. Heavy lines are of considerable use, in the case of line draw- 
ings particularly, in indicating the forms of bodies, as will be seen 
in future examples. 

§ VI.— Notation. 

22. Under the head of Notation, two points are to be consi- 
dered, the manner of indicating the various lines of the diagram, 
and the lettering. As will be seen by examining PI. I., Figs. 1, 2 
— see Ve, eg, &c. — and 5, the visible lines of the object represented 
are indicated by fuU lines ; lines of construction and invisible lines 
of the object, so far as they are shown, are made in dotted lines. 
The intersections of auxiliary planes with the planes of projection, 
called traces, are represented by broken and dotted lines, as at 
P'QP", PI. I., Fig. 16. 

23. Unaccented letters are used to indicate points of the hori- 
zontal projection; and the same with a single accent, denote the 
vertical projections of the same points. A strict adherence to the 
simple rule of lettering the same point with the same letter, wher- 
ever it is shown, will afford a complete key to very complicated 
diagrams, as will be shown as the course proceeds. 

Since projections adequately represent magnitudes, they are 
called the magnitudes themselves. Thus, the point jt?jo' means the 
point whose projections arejt? andjp'; the line ab — a'b' means the 
one whose projections are ab and a'b' . 



FIRST PRINCIPLES. 



24. In the practical applications of projections, "horizontal pro- 
jections" are usually called "plans," and "vertical projections," 
" elevations." 

Before entering upon the study of the subsequent constructions, 
the terms "perpendicular" and " vertical" should be clearly dis- 
tinguished. " Perpendicular" is a relative term, showing that any 
line or surface, to which it is applied, is at right angles to some 
other line or surface. " Vertical" is an absolute term, at any one 
place, and applies to any line or surface at right angles to a level, 
as a water surface. A vertical line, L, is perpendicular to all hori- 
zontal lines which intersect it, but if the entire system of liues thus 
related were inclined, so that all should be oblique, L would still 
be perpendicular to all the rest, though no longer vertical. 

§ VII. — Of the Use of the Method of Projections. 

25. Under this head it is to be noticed, that all drawings are 
made to serve one or the other of two purposes, i.e. they are made 
for use in aiding workmen in the construction of works ; or in 
rendering intelligible, by means of drawings, the real form and size 
of some existing structure ; or else, they are made for ornament, 
or to embellish our houses and gratify our tastes, and to show the 
apparent forms and relative sizes of objects. 

26. Drawings of the former kind are often called, on account of 
the uses to which they are applied, "mechanical" or "working" 
drawings. Those of the latter kind are commonly called pictures ; 
and here it is to be noticed that if " working " drawings are to 
show the true, and not the apparent, proportions of all parts of an 
object, they must, all and always, conform to this one rule, viz. 
All those lines which are equal and similarly situated on the object, 
must be equal and similarly situated on the drawing. 

But, as is now abundantly evident, drawings made according to 
the method of projections, do conform to this rule; hence their use, 
as* above described. 



CHAPTER n. 

PROJECTIONS OF LINES : PROBLEMS IN RIGHT PROJECTION ; AND IN 
OBLIQUE . PROJECTION, SHOWING TWO SIDES OF A SOLID RIGHT 
ANGLE. 

27. Remark. — The style of execution of the problems of this 
chapter is so simple, and so nearly alike for all of them, that it 
need not be described for each problem separately, but will be 
noticed from time to time in connection with groups of figures 
which are similarly executed. 

§ I. — Projections of Straight Lines. 

28. Prob. 1. To construct the projections of a vertical straight 
line, 1^ inches long, whose lowest point is - 1 an inch from the 
horizontal plane, and all of whose points are % of an inch from 
the vertical plane. 

Remarks, a. The remaining figures of PI. I. are drawn just 
half the size indicated by the dimensions given in the text. It 
may be well for the student to make them of full size. 

b. Let MG be understood to be the ground line for all of the 
above problems, without further mention of it. 

1st. Draw, very lightly, an indefinite line perpendicular to the 
ground line, PI. I., Fig. 7. 

2d. Upon it mark a point, a', two inches above the ground line, 
and another point, b ', half an inch above the ground line. 

3 d. Upon the same line, mark the point a,b, three-fourths of an 
inch below the ground line. Then a' b' will be the vertical, and 
ab the horizontal projection of the required line. (8 a) 

29. Prob. 2. To construct the projections of a horizontal line, 
\\ inches long, \\ inches above the horizontal plane, perpendicidar 
to the vertical plane, and toith its furthermost point — fro?n the eye 
— \ of an inch from that plane. PI. I., Fig. 8, in connection with 
the full description of the preceding problem, will afford a sufficient 
explanation of this one. 

Remark. It often happens that a diagram is made more intel- 



10 PROJECTIONS OF LINES. 

ligible by lettering it as at ab, PL I., Fig. 7, and at c'd\ PL I. Fig. 
8, for thus the notation shows unmistakably, that ab or c'd' are not 
the projections of points but of lines. 

30. Problems 3 to 8, inclusive, need now only to be enunciated, 
with references to their constructions, in PI. I. 

Fig. 9 shows the projections of a line, 2j inches long, parallel to 
the ground line, 1-^ inches from the horizontal plane, and 1 inch 
from the vertical plane. 

Fig. 10 is the representation of a line, 2 inches long ; parallel to 
the horizontal plane, and 1 inch above it ; and making an angle of 
30° with the vertical plane. 

Fig. 11 represents a line, 2\ inches long, parallel to the vertical 
plane, and If inches from it, and making an angle of 60° with the 
horizontal plane. 

Fig. 12 gives the projections of a line, 1^ inches long, lying in 
the horizontal plane, parallel to the ground line, and lj inches 
from it. 

Fig. 13 shows the projections of a line, lj inches long, lying in 
the vertical plane, parallel to the ground line, and 1 inch from it. 

Fig. 14 indicates a line, 2-J- inches long, lying in the vertical 
plane, and making an angle of 60° with the horizontal plane. 

31. Prob. 9. To construct the projections of a line which is in a 
plane perpendicular to both planes of projection, the line being 
oblique to both planes of projection. PI. I., Fig. 1 5, shows a pic- 
torial model of this problem. AB represents the line in space ; ab 
its horizontal projection; a'5'its projection on the principal verti- 
cal plane M P'; and A'B' its projection on an auxiliary vertical 
plane P'QP" which is parallel to the given line AB. Since the 
line AB is parallel to this auxiliary plane, its projection upon it — 
A' B'— equals A B. 

32. Now in causing all three of the planes just referred to, to 
coincide with the plane of the paper, taken as a horizontal plane, 
the plane P'QP" is revolved about P'Q, as an axis, till it coincides 
with the vertical plane produced, as at P'QV"; and then the whole 
vertical plane MP V" is revolved backwards into the horizontal 
plane of projection. Now observe, with regard to the first of 
these revolutions, that as the axis P'Q is a vertical line — being the 
intersection of two vertical planes — all points of the arc A V will 
be at the same height above the horizontal plane MP". The same 
is true of the arc B'5", hence a" and a are at equal heights above 



PROJECTIONS OF LINES. 11 

the ground line, and so are b" and b', from which facts we may pass 
at once to the general principle that two or more different elevations 
(vertical projections) (24) of the same point, as A, will be in a line, 
a a", parallel to the ground line, or in space, at equal heights, a't 
and A' a"" above the common horizontal plane MP". 

33. Some other principles may be noted, and remarks made, in 
connection with PI. I., Fig. 15. 

a. A! m being the vertically-projecting line of the point A', ma" 
is the vertical projection of the arc A' a" on the plane QV". a'" a"" 
is evidently the horizontal projection of the same arc. A moment's 
reflection upon these facts will lead at once to the principle, that 
when an arc, or circle, is parallel to either plane of projection, its 
projection on that plane will be an equal arc, or circle, and on the 
other plane, a straight line parallel to the ground line. 

b. a"" b"" maybe called the horizontal projection of the projec- 
tion of AB on the plane P'QP, and it evidently coincides with 
the projection of ab on the same plane. Likewise, mn may be 
called the vertical projection of the projection of AB on the plane 
P'QP, and it is plainly identical with the projection of the projec- 
tion a'b' upon the same plane P'QP. 

c. Inspection of the figure shows that the distance Bb f of any 
point, as B, from the vertical plane of projection equals bt, and 
that bt equals b"n, the distance of the second elevation of B from 
the vertical trace P'Q of the auxiliary vertical plane. 

d. Further inspection shows, that the second elevation shows 
the line in its true length, that the angle made by a"b" with H'q, 
equals the angle made by the line AB with the horizontal plane, 
and that the angle made by a"b" with P'Q equals the angle made 
byAB with the vertical plane. Furthermore, each of these angles 
is evidently the complement of the other. 

34. After this extended preliminary examination of all the facts 
connected with this problem, as shown in PI. I., Fig. 15, every 
point in PL I., Fig. 1 6, will probably be intelligible, as like parts 
of both figures have the same letters. Let the line be 2 inches 
long, and let it make an angle of 60° with the horizontal plane, 
a"b" is the auxiliary elevation on the vertical plane P'QP, after 
that plane has been revolved into the primitive vertical plane, and 
shows the true length and position of the given line. Suppose the 
line in space to be 1|- inches to the left of the auxiliary vertical 
plane P'QP then a b ', its vertical projection, will be perpendicular 
to the ground line, between the parallels a" a! and b"b' (32), and 1^- 
inches from P'Q. The horizontal projection, ab, will be in a'b* 



12 PROJECTIONS OF SOLIDS. 

produced. b"n — b'"b"" are the two projections of the arc in which 
the point b" revolves back to its position, n — b"", in the plane 
P'QP, and b""b — nb' is the line in which rib"" is projected back 
to its primitive position b'b. Therefore, b is at the intersection 
of b""b with a'b' produced, a is similarly found, giving ab as the 
horizontal projection of the given line. 

35. Execution. The foregoing problems are to be inked with 
very black ink; the projections of given lines, and the ground line, 
in heavy full lines; and the lines of construction mfine dotted lines 
as shown in the figures. Lettering is not necessary, except for 
purposes of reference, as in a text book, though it affords occasion 
for practice in making small letters. 

On the other hand, poorly executed lettering disfigures a 
diagram so much that it should be made only after some previous 
practice, and then carefully ; making the letters small, fine, and 
regular. 

§ II. — Hlght Projections of Solids. 

36. Remarh. The term " right projection '' becomes significant 
only when it refers to bodies which are, to a considerable extent, 
bounded by straight lines at right angles to each other. Such 
bodies are said to be drawn in right projection when most of their 
bounding lines are parallel or perpendicular to one or the other 
of the planes of projection. 

37. Peob. 10. — To construct the projections of a vertical right 
prism, having a square base ; standing upon the horizontal plane, 
and with one of its faces parallel to the vertical plane. PI. II., 
Fig. IV. 

Let the prism be 1 inch square, 1^ inches high, and \ of an inch 
from the vertical plane. 

1st. The square ABEF, J of an inch from the ground line, is the 
plan of the prism, and strictly represents its upper base. 

2d. A'BCD', 1£ inches high, is the elevation of the prism, and 
strictly represents its front face. 

3d. The vertical edge A'C, is horizontally projected at A ; and 
so each corner of the plan is the horizontal projection of some one 
of the vertical edges of the prism. 

4th. The vertical face A'B'C'D' has for its horizontal projection 
the line AB ; and so each side of the plan is the horizontal projec- 
tion of one of the vertical faces of the prism. 

§th f The horizontal edge AE, is vertically projected at A 7 ; and 



PROJECTIONS OF SOLIDS. 13 

so each corner of the elevation is the vertical projection of some 
edge which is perpendicular to the vertical plane. 

6th. The vertical projection of the upper base, is A'B'; the 
vertical projection of the lower base, is CD', because that lower 
base is in the horizontal plane of projection. 

1th. The right hand vertical face of the prism, has for its vertical 
projection BT>', and for its horizontal projection BF. 

The style of execution of the figures of this plate is, in general, 
sufficiently indicated by the diagrams. 

38. Pkob. 11. — To construct the plan and two elevations of a 
prism having the proportions of a brick, and placed with its length 
parallel to the ground line. Plate II., Fig. 18. 

1st. abed is the plan, f of an inch broad, twice that distance in 
length, and f of an inch from the ground line, showing that the prism 
in space is at the same distance from the vertical plane of projection. 

2nd. a'b'e'f is the elevation, § of an inch thick, and as long 
as the plan ; and J of an inch above the ground line, showing that 
the prism in space is at this height above the horizontal plane. 

3rd. If a plane, P'QP, be placed perpendicular to both of the 
principal planes of projection, and touching the right hand end 
of the prism, it is evident that the projection of the prism upon 
such a plane will be a rectangle, equal, in length, to the width, bd, 
of the plan, and, in height, to the height, b'f of the side elevation. 
This new projection will also, evidently, be at a distance from the 
primitive vertical plane, i.e. from P'Q, equal to c?Q, and at a dis- 
tance from the horizontal plane equal to Qf. When, therefore, 
the auxiliary plane, P'QP, is revolved about P'Q into the primitive 
vertical plane of projection, the new projection will appear at 
a"e"c"g". 

4th. dc'" is the horizontal, and b'c" the vertical projection of 
the arc in which the point db' revolves into the primitive vertical 
plane. ba"\ b'a", are the two projections of the horizontal arc in 
which the corner bb' of the prism revolves. 

39. Peob. 12. — To construct the two projections of a cylinder 
ichich stands upon the horizontal plane. PI. II., Fig. 19. 

The circle AciBb is evidently the plan of such a cylinder, and 
the rectangle A'B'CT) its elevation. Observe, here, that while the 
elevation, alone, is the same as that of a prism of the same height, 
Fig. 17, the plan shows the body represented, to be a cylinder. 

40. As regards execution, the right hand line B'D' of a cylinder 



14 PROJECTIONS OF SOLIDS. 

or cone may be made less heavy than the line B'D', Fig. 17 ; and 
in the plan, the semicircle, aBb, convex towards the ground line, 
and limited by a diameter ab, which makes an ongle of 45° with 
the ground line, is made heavy, but gradually tapered, into a fine 
line in the vicinity of the points a and b. 

41. Prob. 13. — To construct the projections of a cylinder whose 
axis is placed parallel to the ground line. PI. II., Fig. 20. 

Let the cylinder be 1^ inches long, f of an inch in diameter, its 
axis § of an inch from the horizontal plane, and ^ an inch from the 
vertical plane. The principal projections will, of course, be two 
equal rectangles, gehf and a'b'c'd ', since all the diameters of 
the cylinder are equal. The centre lines, g'h' and ab, are made at 
the same distances from the ground line, that the axis of the cylin- 
der is from the planes of projection. 

42. Taking now into consideration the mere form of these two 
projections, they could not be distinguished from the projections of a 
square-based prism of equal length and diameter. There are, how- 
ever, three ways of distinguishing between the projections of a hori- 
zontal cylinder and of a horizontal prism having, a square base. 
.First: In the case of the cylinder, the lines c'd' and ef are made 
quite moderately heavy, rather than fully heavy, as they would be 
on the projections of a prism. Second: The lettering shows plainly, 
at a glance, that PI. II., Fig. 20,-represents a cylinder ; thus, g'h' is 
the vertical projection of three lines; viz., the element gh, which is 
furthest from the vertical plane, ab, considered as the horizontal 
projection of the axis, and ef, the element nearest to the vertical 
plane. Again, ab is the horizontal projection of three lines ; viz., of 
a'b' the highest element, of g'h' the axis, and of c'd', the lowest 
element. Third: An end elevation shows, that in PI. II., Fig. 20, 
a cylinder is represented. In the figure, the auxiliary vertical plane 
P'QP does not coincide with the base of the cylinder, but is at a dis- 
tance, ef", from it, and is revolved towards the left, to coincide 
with the primitive vertical plane of projection. Excepting this, 
the construction is so similar to that of Prob. 11, as to need no 
farther explanation. 

In inking, the end elevation, b"f"d\ is made heavy at nf"d"p, 
and tapered into a fine line in the vicinity of n andjt?. 

§ 111.— Oblique Projections shoioing tico sides of a Solid Bight 

Angle. 

43. A solid right angle is an angle such as that at any corner of 



PROJECTIONS OF SOLIDS. 15 

a cube, or like the space in the corner of a square room, and 
is bounded by the three plane right angles which form such 
corners. 

When a cube, for example, is held so that two of its vertical 
faces can be seen, we should not look at them directly or in a 
direction perpendicular to either of them, and the cube would be 
said to be viewed obliquely, and where drawn in this position, 
would be said to be represented in oblique projection. 

44. Peob. 14. — To construct the plan and two elevations of a 
vertical prism, with a square base ; resting on the horizontal plane, 
and having its vertical faces inclined to the vertical plane of pro- 
jection. PL II., Figs. 21-22. 

1st. ABCG is the plan, with its sides placed at any convenient 
angle with the ground line. 

2c?. A'B'D'E' is the vertical projection of that vertical face 
whose horizontal projection is AB. 

3d. B'C'E'F' is the vertical projection of that face whose hori- 
zontal projection is BC. This completes the vertical projection 
of the visible parts of the prism, when we look at the prism in the 
direction of the lines CF', &c. 

4th. Let gb be the horizontal trace of an auxiliary vertical plane 
of projection, which is perpendicular to both of the principal planes 
of projection. In looking perpendicularly towards this plane, i.e. 
in the directions Gg, &c, AG and AB are evidently the horizontal 
projections of those vertical faces that would then be visible; and 
the projecting lines, Gg, Aa, and Bb determine the widths ga and 
ab of those faces as seen in the new elevation. Now the auxiliary 
plane gb is not necessarily revolved about its vertical trace (not 
shown), but may just as well be taken up and transferred to any 
position where it will coincide with the primitive vertical plane ; 
only its ground line gb must be made to coincide with the principal 
ground line, as at H'E". Hence, making H'D" and D "E" respec- 
tively equal to ga and ab, and by drawing WG", <fcc, the new 
elevation will be completed. 

45. By inspecting the two elevations — PI. II., Figs. 21-22 — it 
appears that they are identical in form, but that the parts of Fig. 
22 are not identical with those of Fig. 21. 

To prevent the student from constructing this problem merely 
by rote, the same faces on the two elevations are distinguished by 
marks. Thus the surfaces marked Xt are the two elevations of the 
same face of the prism ; the. one marked q> is visible only on the 



16 PROJECTIONS OF SOLIDS. 

first elevation, and the one marked x is visible only on the second 
elevation — Fig. 22. 

46. PI. II., Fig. 23, represents a small quadrangular prism in two 
elevations, the axis being horizontal in space, so that the left hand 
elevation shows the base of the prism. In the practical applica- 
tions of this construction, the centre, s, of the square projection is 
generally on a given line, not parallel to the sides of the square. 
Hence this construction affords occasion for an application of the 
problem : To draw a square of given size, with its centre on a given 
line, and its sides not parallel to that line. The following solution 
should be carefully remembered, it being of frequent application. 
Through the given centre, s, draw a line, L, in any direction, and 
another, L', also through s, at right angles to L. On each of these 
lines, lay off each *way from s, half the length of a side of the 
square. Through the points thus formed, draw lines parallel to 
the lines L and L' and they will form the required square whose 
centre is s. 

47. Peob. 15. — To construct the plan and several elevations of a 
vertical hexagonal prism, which rests upon the horizontal plane of 
projection. PI. II., Figs. 24, 25, 26. 

Remarh. The distinction between right and oblique projection 
becomes obscure as we pass from the consideration of bodies whose 
surfaces are at right angles to each other. Figs. 24 and 25 show a 
hexagonal prism as much in right projection as such a body can be 
thus shown, but, as in both cases a majority of its surfaces are, consi- 
dered separately, in oblique projection, its construction is given here. 

48. In Fig. 24 the hexagonal prism is, as shown by the plan, 
placed so that two of its vertical faces are parallel to the vertical 
plane of projection. Observe that where the hexagon is thus 
placed, three of its faces will be visible, one of them in its real size, 
viz,. BC, B'C'F'G', and that the extreme width, E'H', of the ele- 
vation, equals the diameter of the circumscribing circle of the plan. 
Notice, also, that as BC equals half of AD, while AB and CD are 
equal, and equally inclined to the vertical plane, the elevations, A'F' 
and CD', of these latter faces, will be equal, and each half as wide 
as the middle face. This fact enables us to construct the elevation 
of a hexagonal prism situated as here described, without construct- 
ing the plan, provided we know the width and height of one face 
of the prism. This last construction should be remembered, it 
being of frequent and convenient application in the drawing of 
nuts, bolt-heads, &c, in machine drawing. 



PEOJECTIONS OF SOLIDS. 



17 



49. PI. II., Fig. 25 shows the elevation of the same prism on a 
plane which' origin ally was placed at ib, and perpendicular to the 
horizontal plane ; whence it appears, that if a certain elevation of a 
hexagonal prism shows three of its faces, and one of them m its 
full size, another elevation, at right angles to this one, will show 
but two faces, neither of them in its full size ; the extreme width, 
I"B", of the second elevation being equal to the diameter of the 
inscribed circle of the plan. 

50. PL II., Fig. 26 shows the elevation of the same prism as it 
would appear if projected upon a vertical plane having the position 
indicated by the horizontal trace jb", when that plane had been 
transferred to the principal vertical plane at Fig. 26. In this ele- 
vation, none of the faces of the prism are seen in their true 
size. The auxiliary vertical plane, whose horizontal trace is 
jb", could have been revolved about that trace, directly back 
into the horizontal plane, causing the corresponding elevation 
to appear in the lines T>d, &c, produced to the left of jb" as 
a ground line. 

Elevations on auxiliary vertical planes can always be made to 
appear thus, but it seems more natural to have the several eleva- 
tions side by side above the principal ground line— a result which 
is accomplished by transferring the auxiliary planes as heretofore 
described. Fig. 27 represents two elevations of a hexagonal prism, 
placed so as to show the base in one elevation, and three of its 
faces, unequally, in the other. The centre of the elevation which 
shows the base, may be made in a given line perpendicular to o'g\ 
by placing the centre of the circumscribing circle used in con- 
structing the hexagon, upon such a line. Having constructed this 
elevation, project its points, a,b, &c, across to the other vertical 
plane, P', which is in space perpendicular to the plane, P, at the 
line, o y. By representing the elevation on P' as touching o'g', we 
indicate that the prism touches the plane, P, just as the elevation 
in Fig. 24, indicates that the prism there shown rests upon the 
horizontal plane. 

51. Peob. 16. — To construct the plan and two elevations of a 
pile of blocks of equal widths, but of different lengths, so placed 
as to form a symmetrical body of uniform width. PL III., 

Fig. 28. 

The auxiliary vertical plane of projection, perpendicular to the 
horizontal plane at &'"/"', is made to coincide with the principal 
vertical plane by direct revolution. The point a'"a"", the projec- 

2 



18 PROJECTIONS OF SOLIDS. 

tion of aa' on the auxiliary vertical plane, revolves in a horizontal 
arc, of which a'"a" is the horizontal, and a"" a" the vertical pro- 
jection (33 a), giving a", a point of the second elevation. Other 
points of this elevation are found in the same way. This figure 
differs from Figs. 18 and 20, of Plate II., only in presenting more 
points to be constructed. If the student finds any difficulty with 
this example, let him refer to those just mentioned, and to first 
principles. 

52. Pkob. 17. To construct the oblique projection of a vertical 
circle. PL III., Fig. 30. 

Let the circle first be placed parallel to the vertical plane of pro- 
jection, and tangent to the horizontal plane. Its vertical projection 
will then be a circle a'b'e'g', tangent to the ground line, and its 
horizontal projection will be a straight line, parallel to the ground 
line, and equal to the diameter of the circle. Now let a vertical 
line, F — Ff, be drawn tangent to the circle at FF', and let the 
circle be revolved about this line as an axis — still being kept ver- 
tical — till it makes a certain angle with the vertical plane, as shown 
by the angle which the new horizontal projection, FB, makes with 
the ground line. Now, in this revolution, any point, b'b, of the 
primitive position of the circle, describes a horizontal arc in space, 
of which #B is the horizontal projection — equal to the arc in space 
(33 a) — and #'B' is the vertical projection, giving B' a point in the 
new elevation. Other points, as c'c, a r a, which, being in a vertical 
line, are horizontally projected in the same point, c — «, revolve in 
horizontal arcs, of which c — a; A — C is the horizontal projection, 
and a' A' c'C; are the vertical projections, giving A' and C, other 
points of the new elevation. The remaining points of the new 
elevation are found in a manner precisely similar, as is fully shown 
in the figure. This being a plane problem, no part of it need be 
hiked in heavy lines. 

53. Peob. 18. — To construct the projections of a cylinder ichose 
convex surface rests on the horizontal plane, and whose axis is 
inclined to the vertical plane. PL III., Fig. 31. 

As may be learned by reflecting upon Fig. 19, of Plate II., the 
principle is general that the projection of a right cylinder with a 
circular base, upon any plane to which its axis is parallel, will be a 
rectangle. Therefore let CSTV, PL III., Fig. 31, be the plan of 
the cylinder. Since it rests upon the horizontal plane, q'u\ in the 
ground line, is the vertical projection of its line of contact with that 



PROJECTIONS OF SOLIDS. 19 

plane, an&p'A' is the vertical projection of pA, the highest element 
of the cylinder, as it is at a height above the ground line, equal to 
the diameter, TV, of the cylinder. The construction of the verti- 
cal projection of either base, is but a repetition of the last problem. 
In the figure, the left hand base is found just as in the preceding 
problem, and the construction, being fully given, needs no further 
explanation. 

54. The vertical projection of the base TV is found somewhat 
differently. TV may properly be considered as the horizontal 
diameter of this base. Then let the base be revolved about TV, 
till it becomes parallel to the horizontal plane of projection. It 
will then appear as a circle, and a line, as n"n, will show the true 
height of n above the diameter TV. So, also, o"o will show the 
true distance of o below TV. Therefore the vertical projections 
of the points n and o, will be in the line n — n', perpendicular to the 
ground line, and at distances above and below TV, the vertical 
projection of TV, equal, respectively, to nn" and oo" . Having, in 
the same manner, found r' and t\ the vertical projections of two 
points whose common horizontal projection t — r is assumed, as 
was n — o, the vertical projection of the base TV can be drawn by 
the help of the irregular curved ruler. 

55. In the execution of this figure, SV is made slightly heavy, 
and TV fully heavy, and, for reasons w T hich cannot here be fully 
explained, the portion, n'T't', of the elevation of the right hand 
base, and the small portion, DV, of the left hand base, are made 
heavy. Suffice it to say : First. That a part of the convex surface 
is in the light, while the right hand base is in the dark. Second. 
n'T't' divides the illuminated half of the convex surface, from the 
base at the right, which is in the dark ; and D V divides the illu- 
minated left hand base from the visible portion of the darkened 
half of the convex surface (18-20). 

56. Prob. 19. To construct the two projections of a right cone, 
with a circular base in the horizontal plane ; and to construct 
either projection of a line, drawn from the vertex to the circum- 
ference of the base, having the other projection of the same line 
given. PL III., Fig. 32. 

Remark. When the axis of a cone is vertical, perpendicular to 
the vertical plane, or parallel to the ground line, the cone is shown 
in right projection as much as such a body can be, but as all the 
straight lines upon its surface are then inclined to one or both 
planes of projection, the above problem is inserted here among 
problems of oblique projections. 



20 PROJECTIONS OF SOLIDS. 

57. Let VB be the radius of the circle, which, with the point V, 
is the horizontal projection of the cone. Since the base of the cone 
rests in the horizontal plane of projection, C B' is its vertical 
projection. Since the axis of the cone is vertical, Y', the vertical 
projection of the vertex, must be in a perpendicular to the ground 
line, through V, and may be assumed, unless the height of the 
cone is given. V'C and VB', the extreme elements, as seen in 
elevation, are parallel to the vertical plane of projection, hence 
their horizontal projections are CY and BY, parallel to the ground 
line (8 e). Let it be required to find the horizontal projection of 
any element, whose vertical projection, YD', is given. Y is the 
horizontal projection of Y', and D', being in the circumference oi 
the base, is horizontally projected at D, therefore YD is the hori- 
zontal projection of that straight line in the conic surface, whose 
vertical projection is YD'. Having given, YA, the horizontal 
projection of a straight line in the surface of the cone, let it be 
required to find its vertical projection. Y' is the vertical projec- 
tion of Y, and A, being in the circumference of the base, is verti 
cally projected at A'. Therefore V'A' is the required vertical pro- 
jection of the proposed line. In inking the figure, no part of the 
plan is heavy lined, and in the elevation, only the element Y B' is 
slightly heavy. 

58. Peob. 20. To construct the projections of a right hexagonal 
prism / whose axis is oblique to the horizontal plane, and parallel 
to the vertical plane. PI. III., Figs. 33, 34. 

1st. Commence by constructing the projections of the same prism 
as seen when standing vertically, as in Fig. 33. The plan only is 
strictly needed, but the elevation may as well be added here, for 
completeness' sake, and because some use can be made of it. 

2nd. Draw J"G", making any convenient angle with the ground 
line, and set off upon it spaces equal to G'J', JTT, and JT, from 
Fig. 33. 

3rd. Since the prism is a right one, at J", &c, draw perpen- 
diculars to J /; G", make each of them equal to J'C, Fig. 33, and 
draw F^C", which will be parallel to J"G", and will complete the 
second elevation. 

4th. Let us suppose that the prism was moved from its first 
position, Fig. 33, parallel to the vertical plane, and towards the 
right, and then inclined, as described, with the corner, CJ', of the 
base, remaining in the horizontal plane. It is clear that all points 
of the new plan, as B'", would be in parallels, as BB"', to the 



PROJECTIONS OF SOLIDS. 21 

ground line, through the primitive plans, as B, of the same points. 
It is equally true that the points of the new plan will be in perpen- 
diculars to the ground line through the new elevations B", &c, of 
the same points (15), hence these points B'", &c, will be at the 
intersections of these two groups of lines. Thus, A'" is at the 
intersection of AA'" with A" A'" ; C" is at the intersection of CC" ; 
with C"C"; K'" is at the intersection of DK'" with H"K'", &c. 

5th. B'"C'", F'"E'", and G'"K'", being the projections of lines 
of the prism which are parallel in space, are themselves parallel. A 
similar remark applies to C'"D'", A"'F'", and H'"G'". Observe, 
that as the upper or visible base is viewed obliquely, it is not seen 
in its true size, F"'C" being less than FC, Fig. 33; so that this 
base A'"C", E'", does not appear in the new plan as a regular 
hexagon. 

59. Pkob. 21. To construct the projections of the prism, given 

in the previous problem, when its edges are inclined to both 

planes of projection. PI. III., Fig. 34a. 

Remark. This problem involves the showing of three faces of 
a solid angle, but, with a similar one respecting the pyramid, is 
retained here, on account of its usefulness as a study, in giving 
completeness to the general problem of the projection of the prism. 

If the prism, PL III., Fig. 34, be moved to any new position, 
such that the inclination of its edges to the vertical plane, only, 
shall be changed, the inclination of its edges to the horizontal 
plane of projection being unchanged, the new'plan will be merely 
a copy of the second plan, placed in a new position. Let the par- 
ticular position chosen be such that the axis of the prism shall be 
in a plane perpendicular to the ground line, i.e. to both planes of 
projection ; then the axis of symmetry, C'"G'", of the second 
plan, will take the position C""G'"', and on each side of this line the 
plan, Fig. 34 a, will be made, similar to the halves of the j)lan in 
Fig. 34. 

As the prism is turned horizontally about the corner J", and 
then transferred, producing the result that the inclination of its 
axis to the horizontal plane is unchanged, all points of the third 
elevation, as A'"", 0'"", &c, will be in parallels to the ground line 
through A", C", &c, and in perpendiculars to the ground line, 
through A"", C"", &c. 

60. By examination of this solution, and by inspection of Figs. 
34 and 34a, it appears that a change in the position of the axis, 
with reference to but one plane of projection at a time, can be 



22 PROJECTIONS OF SOLIDS. 

represented directly from projections already given; also that a 
curve, beginning with the first plan, and traced through the six 
figures composing the three given pairs of projections in the order 
in which they must be made, would be an S curve, ending in the 
third elevation. 

61. Execution. — The full explanation of the location of the heavy 
lines cannot here be given. The careful inquirer may be able to 
satisfy himself that the heavy lines of the figures, as shown, are the 
projections of those edges of the prism which divide its illuminated 
from its dark surfaces. 

62. Peob. 22. To construct the projections of a regular hexa- 
gonal pyramid, whose axis is inclined to the horizontal plane only. 
PI. III., Figs. 35, 36. 

1st. Commence, as with the prism in the last problem, by repre- 
senting the pyramid as having its axis vertical. 

2nd. Draw a"d", equal to a'd', and divided in the same way. At 
»", the middle point of «"<?", draw n"Y" perpendicular to a"d", 
and make it equal to n'V, which gives V" the new elevation of the 
vertex. Join V" with a", b' ', c", and d" , and the new elevation 
will be completed. 

3rd. Supposing the same translation and rotation to occur to the 
primitive position of the pyramid, that was made in the case of the 
prism (52, 4th), the points of the new plan, Fig. 36, will be found 
in a manner similar to that shown in Fig. 34. Y" f is at the inter- 
section of W" with V"V" ; c'" is at the intersection cc'" with 
c"c'" \ d'" is at the intersection of dd'" with d"d"\ &c. 

4th. The points, a"'b" r c'" .... /'"; of the base, are connected 
with V", the new horizontal projection of the vertex, to complete 
the new plan. If the pyramid were less inclined, the perpendicular 
V"V" would fall within the base, and the whole base would then 
be visible in the plan. As it is, f'"a'" and a"'b'" are hidden, and 
therefore dotted. 

5 th. The heavy lines are correctly placed in the diagram ; also 
the partially heavy lines, which are all between Y'"d r " and the 
ground line, but the reasons for their location cannot here be given, 
beyond the general principle (18-20) already given. 

63. Peob. 23. To construct the projections of the regular hexa- 
gonal pyramid, when its axis is oblique to both planes of projec- 
tion. PI. III., Fig. 36a. 

Suppose the pyramid here shown to be the one represented in 



ELEMENTARY INTERSECTIONS. 23 

figures 35 and 36, and suppose that it has been turned horizontally 
about the corner, a!' a'", Fig. 36, of the base, and then transferred 
to the position represented in Fig. 36<2. After the pyramid has 
been thus moved, the third plan will be merely a copy of the second 
plan, placed so that its axis of symmetry, Y""d"", shall make any 
assumed angle with the ground line. Make the distances on 
Y""d" n equal to those on Y'"d"\ and let c""e n " and b""f ! " equal 
c'"e'" and b"f", after which the third plan is completed by join- 
ing the points of the base with V"", the vertex. 

Next we have to consider, that as the inclination of the axis of 
the pyramid to the horizontal plane of projection is unchanged, the 
points, as Y"'", of the third elevation, will be at the intersection of 
parallels to the ground line, through the corresponding points, as 
V", of the second elevation, with perpendiculars through the same 
points, as V"", seen in the third plan. Observe that the two points 
vertically projected in c" , being at the same height above the 
ground line, will appear in the third elevation at "'" and e""\ in 
the same straight line, through c", and parallel to the ground line. 

Remembering also that lines which are parallel in space must 
have parallel projections, on the same plane, c"'"d"'" will be paral- 
lel tof""a""\ &c. 

The heavy lines are indicated in the figure. 

§ IV. — Special Elementary Intersections and Developments. 

64 (1). The positions of other planes, than those of projection, 
are indicated by their intersections with the planes of projection. 
These intersections are called traces. 

A plane can cut a straight line in only one point ; hence, if a 
plane cuts the ground line at a certain point, its traces, both being 
in the plane, must meet in that point. 

In PI. I., Fig. 5, bBb'k' is a plane perpendicidar to the ground 
line, MG, and, therefore, to both planes of projection, and we see 
that its two traces, bh' and b h' , are perpendicular to the ground 
line at k'. Likewise in PL L, Fig. 15, Aaa't is a plane perpendicu- 
lar to the ground line MQ, and its traces at and at are perpendi- 
cular to MQ. That is: if a plane is perpendicular to the ground 
line, its traces will also be perpendicidar to that line. 

This is seen in regular projection, in PI. I., Fig. 16, where PQ is 
the horizontal trace, and P'Q, the vertical trace, of such a plane. 

In PI. I., Fig. 5, ¥I\fk is a plane, parallel to the vertical plane, 
and it has only a horizontal trace, fJc, which is parcdlel to the 
ground line. The same is true for all such planes. Likewise, 



24 ELEMENTARY INTERSECTIONS. 

ABa'5' is a horizontal plane. All such, planes have only a vertical 
trace, as a'b', parallel to the ground line. 

In PL I., Fig. 2, the plane Yv'd'd is perpendicular only to the 
vertical plane, and, as the figure shows, the horizontal trace only, 
as dd', of such a plane, is perpendicular to the ground line. Also 
the angle v'd'b', between the vertical trace, v d', and the ground 
line, is the angle made by the plane with the horizontal plane. 

In like manner, it can easily be seen that, if a plane be perpen- 
dicular only to the horizontal plane, as in case of a partly open door, 
its vertical trace only (the edge of the door at the hinges) will be 
perpendicular to the ground line, and the angle between its hori- 
zontal trace and the ground line, will be the angle made by the 
plane with the vertical plane of projection. 

Finally, if a plane is oblique to both planes of projection, both 
of its traces will be oblique to the ground line, and at the same 
point. Thus, PI. I., Fig. 6, may represent such a plane, having LF 
for its horizontal, and L'F for its vertical trace. 

All the principles just stated can be simply illustrated by taking 
a book, half open, for the planes of projection, and either of the 
triangles for the given movable plane ; and when clearly under- 
stood, the following problems can also be easily comprehended. 

Peob. 24. — To find the curve of intersection of a cylinder with 
a plane. PI. III. A., Fig. 1. 

Let the cylinder, ADBG— A'B", be vertical, and the cutting 
plane, PQP', be perpendicular only to the vertical plane. All 
points in such a plane must have their vertical projections (that is, 
must be vertically projected) in the vertical trace, QP', of the 
plane, but the required curve must also be embraced by the visible 
limits, A'A" and B'B", of the cylinder. Hence, a'b' is the verti- 
cal projection of this curve. Again, as the cylinder is vertical, all 
points on its convex surface must be horizontally projected in 
ADBG. Hence, this circle is the horizontal projection of the 
required curve. 

64 (2). When a plane revolves about any line in it as an axis, 
every point of it, not in the axis, moves in a circular arc, whose 
radii are all perpendicular to the axis. The representation of the 
revolution is much simplified by taking the axis in, parallel to, or 
perpendicular to, a plane of projection. 

Prob. 25. — To revolve the curve found in the last problem, so as 
to show its true size. 



PL.I. 




ELEMENTARY INTERSECTIONS. 25 

Let AB — a'b', the longer axis of the curve, and which is parallel 
to the vertical plane of projection, be taken as the axis of revolu- 
tion. The curve may then be revolved till parallel to that plane, 
when its real size and form will appear. Then, at c', d\ <fcc, the 
vertical projections of C and H, D and G, &c, draw perpendicu- 
lars, as c"h", to a'b', and make c'c" ' =c'h" '=nC Proceed likewise 
at d', &c, since the lines, as nC, being parallel to the horizontal 
plane, are seen in their true size in horizontal projection ; and join 
the points a', h',g', &c, which will give the required true form and 
size of the curve of intersection before found. 

Remark. — This curve is an oval, called an ellipse. Its true size 
could have been shown by revolving its original position about 
DG as an axis, till parallel to the horizontal plane. The student 
may add this construction to the plate. 

64 (3). The convex surface of a cylinder is wholly composed of 
straight lines, called elements, parallel to its axis. The convex sur- 
face of a cone is composed of similar elements, all of which meet 
at its vertex. Hence, each of these surfaces can evidently be 
rolled upon a plane, till the element first placed in contact with the 
plane, returns into it again. The figure, thus rolled over on the 
plane, is called the development of the given convex surface, and 
its area equals the area of that surface. 

Prob. 26. — To develops the portion of the cylinder, PI. III. A., 
Fig. 1, below the cutting plane, PQP'. 

Suppose the cylinder to be hollow as if made of tin, and to be 
cut open along the element B'b'. Then suppose the element A! a' 
to be placed on the paper, as at AV, Fig. 2, and let each half be 
rolled out upon the paper. The part ADB will appear to the left 
of A' a', and the part AGB, to the right. The base being a circle, 
perpendicular to the elements, will develope into a straight line 
B'B", Fig. 2, found by making A'c = AC, Fig. 1, cd=CD, Fig. 1, 
&c, and A A=AH, Fig. 1, &c. B'B" may also, for convenience, 
be AB', Fig. 1, produced. Then the parallels to AV, through c, 
d, &c, will be developments of elements standing on C, D, &c, 
Fig. 1, and by projecting over upon them, a' at a', c' at c' and h' ; 

d' at d r and g', B' at b' and b' ', and joining the points, the 

figure WWb"a'b', will be the required development of the cylinder. 

Remark. — If, now, a flat sheet of metal be cut to the pattern 
just found, it will roll up into a cylinder, cut off obliquely as by 
the plane PQP'. By making the angle P'QA' of any desired size, 
the corresponding flat pattern can be made as now explained. 



26 ELEMENTARY INTERSECTIONS. 

Prob. 27. — To find the intersection of a vertical cone, with a 
plane, perpendicular to the vertical plane of projection. PI. III. 
A, Fig. 3. 

Let V — ADBG be the plan, and A'B'Y' the elevation of the 
cone, and PQP' the cutting plane. For the reasons given in Prob. 
24, a'b' will be the vertical projection of the required curve. The 
surface of the cone, not being vertical, the horizontal projection of 
the intersection requires operations additional to those of Prob. 24, 
as follows : EE'Y', for example, is an auxiliary plane, perpendicu- 
lar to the vertical plane, and cutting the cone's surface, from the 
vertex to the base, in two elements. As this plane is perpendicular 
to the vertical plane, these elements will be one behind the other, 
and both will be vertically projected in V'E', its vertical trace. 
Their horizontal projections are YE and VF. Again, as the cut- 
ting plane PQP' is perpendicular to the vertical plane, all points in 
it will be vertically projected in its vertical trace P'Q ; hence, e' is 
the vertical projection of two points of the curve, one on the ele- 
ment VE— YE", and the other on YF— Y'E'. Their horizontal 
projections, e and/, are found by projecting down e' , at e, on YE, 
and at /on YF. 

Other points being found in precisely the same manner, thus 
minutely explained, they are left for the student to construct. One 
or two peculiar points, only, need be further noticed. The projecting 
line from d', for instance, coincides with the elements YD and YG, 
in projection ; hence, the horizontal projections of the two points 
represented by d' , must be found otherwise. The following 
method is convenient, and applicable to all other points also. Let 
M'N' be the vertical trace of a horizontal auxiliary plane through 
d'. This plane will cut from the cone the circle ni'n — dmg, on 
which d' can be projected at d and g, the points desired. Again, 
on the cylinder, d', the middle of a'b' is on the axis O — dd '. That 
is, the centre of the ellipse cut from a cylinder, is on the axis of the 
cylinder. Not so, however, with the cone ; p', the middle of a'b'-, 
is not on Y'D', the vertical projection of the axis, but is on the 
side of it towards the lowest point, bb', of the curve of intersec- 
tion. On account of the acuteness of the intersections at^> and q, 
these points can better be found as were d and g. 

The horizontal projections of a! and b ', on the extreme elements, 
are at a and b. 

Remarks. — a. The curve adbg — a'b' is an ellipse, wiiose longer 
axis is the line ab — a'b', whose true length is a'b' ; and whose 
shorter axis is the line j^ — p\ whose true length is pq. 



ELEMENTARY INTERSECTIONS. 27 

b. The true size of the curve can be found by either of the ways 
indicated in Prob. 25, also by revolving the plane PQP', contain- 
ing it, either, about PQ as an axis, into the horizontal plane ; or, 
about P'Q as an axis, into the vertical plane. In the former case, 
it is only to be remembered that e'Q, for example, shows the true 
distance of ee' from PQ; and, in the latter case, that ek, for exam- 
ple, shows the true distance of ee' from the vertical trace P'Q (6). 

Prop,. 28.— To develope the convex surface of a cone, PI. III. 
A., Fig. 4, together with the curve of intersection, found in the last 
problem. 

First. If the element VB— V'B' be placed in contact with the 
paper at V'B', and if the cone be then rolled upon the paper till 
this element returns into it again, as at V'B", the development, 
V'B'B", will be made. As all the elements are equal, and as the 
vertex is stationary, the development of the base will be the arc 
B'B" with a radius equal to V'B', the cone's slant height, and of a 
length equal to the circumference ADBG-. This length is found, 
as in case of the cylinder, by taking equal arcs of the base, so 
small that their chords shall be sensibly equal to them, and laying 
off those chords from B', on the arc B'B", till B" is located. 
Thus, BE being one eighth of ADBG, its length is laid off as at 
Be" eight times to find B". 

Second. To show the curve, adbg—a'V, on the development, 
consider that only the extreme elements, as VB— V'B', show their 
true length in projection. Hence, the points between a' and b' 
must be revolved around the axis of the cone, into these elements, 
in order to show their true distances from the vertex. This axis 
being vertical, the arcs of revolution will be horizontal, and will 
therefore be vertically projected in the horizontal lines c'u, d'n, &c, 
and Vu, Vn, &c, will be the true distances of c\ d\ &c, from the 
vertex. Hence, make W= W; W, and VV=V'«; W, and 
Y' n " =Vn, <fcc, and the curve b'a'b'' will be the development of 
the intersection of the plane PQP' with the cone. 

JZemarlc.— -The remarks made upon the development of the 
cylinder equally apply here. 

Prob. 29.— To find the intersection of a vertical cylinder with 
two horizontal ones ; their axes being in a plane parallel to the ver- 
tical plane of projection. PI. III. A, Fig. 5. # 

ABE— A'B'A'B" is the vertical cylinder, and MNQR— O P O 
P" the lower horizontal cylinder. 



28 ELEMENTARY INTERSECTIONS. 

First. To find the highest and lowest, and foremost and hind- 
most points of the intersection. Since the horizontal cylinder is 
the smaller one, it will enter the vertical cylinder on one side, and 
leave it on the other, giving two curves ; but as one cylinder is ver- 
tical, and the intersection, being common to both, is on it, the 
horizontal projections of both curves are known at once to be 
CAE and DBF. Now A is the horizontal projection of both the 
highest and lowest points of CAE. Their vertical projections are 
a" and a'. Also C and E are the horizontal projections of the fore- 
most and hindmost points, and c', on M'N', midway between O 
and O", is the vertical projection of both of them. (42.) 

In like manner b\ b" and d' are found. 

Second. To find other intermediate points. Take the tw T o points 
whose horizontal projection is G, for example. They are on the 
horizontal elements, one on the upper, and the other on the lower 
half of the horizontal cylinder, and whose horizontal projection is 
ST. But to find their vertical projections, we must revolve one of 
the bases, as MQ, till parallel to a plane of projection. Let this 
base revolve about its vertical diameter, O — O'Q", till parallel to 
the vertical plane, when OM" — 0'M'"0" will be the vertical pro- 
jection of its front half. In this revolution the points, S, revolve 
to S", and will thence be vertically projected at U' and S'. In 
counter revolution, these points return in horizontal arcs to u' and 
s', and u'v' and s't' are the vertical projections of the elements ST. 
Hence, project G, and also H, at g' and g\ h! and A", and we shall 
have four more points of intersection. Any number of points can 
be similarly found. 

Remarks. — a. The last four points could as easily have been 
found, by revolving the base MQ about the horizontal diameter 
MQ — M', till parallel to the horizontal plane. This construction 
is left for the student. 

b. If the axes did not intersect each other, as at II', the points C 
and E would not be equidistant from OP, and would not have one 
point, c', for their vertical projection, and the vertical projection 
of the back half, as AE, of each curve would be a dotted line, 
separate from the same projection of the front half. The student 
may construct this case, also that where one of the elements, MN", 
or QK, does not intersect the vertical cylinder. 

c. The horizontal cylinder, Fig. 6, shows that when the two cylin- 
ders, placed as before, are of equal diameter, the vertical projec- 
tions of their curves of intersection are straight lines. Hence, 
each of the curves themselves is contained in a plane, that is, it is 



ELEMENTARY INTERSECTIONS. 29 

a "plane curve." This figure, if regarded separately, as a plan 
view, therefore may represent the plan of the intersection of two 
equal semi-circular arches, and the curves, KL and AY, of inter- 
section, will be ellipses. 

The curves on the cylinders in Fig. 5 cannot be contained in 
planes. Such curves are said to be of double curvature. 

cl. By developing the cylinders, in Figs. 4 and 5, as in Fig. 2, the 
patterns may be found which will give intersecting sheet metal 
pipes, when rolled up in cylindrical form. The student should 
construct these developments, also the case in which the vertical 
cylinder should be the smaller one. 



Peob. 30. — To find the intersection of a horizontal cylinder with 
a verticcd cone. PI. III. A, Fig. 7. 

Let ABV be the vertical projection of a cone, and let the circle 
with radius oa, be an end view of the cylinder ; its axis, o V, in- 
tersecting A'Y', that of the cone. Let PQ be the vertical trace 
of a second vertical plane, perpendicular to the ground line, as 
in PI. I., Figs. 15 and 16, and let Y'E'D be the vertical projection 
of the cone, and G'G'N'jST" that of the cylinder, on this plane. 
In this construction, therefore, two vertical projections are em- 
ployed, instead of a horizontal and vertical projection, for any two 
projections of an object are enough to show its form and position. 
This will more readily appear by turning the plate to bring YAB 
below PQ, when PQ w T ill be the ground line, the right hand pro- 
jection a plan, and the left hand one an elevation, like Fig. 5. 

Now to find the intersection. Speaking as if facing the vertical 
plane of projection, represented by the paper to the left of PQ, 
after revolving that plane about PQ into the paper, AY — A'Y' is 
the foremost element, and a' is found by projecting a across upon 
A'Y'. Next, DY is the right hand projection of two elements, 
whose left hand projections are EY andDY. We therefore project 
G at g' and e f . 

To find intermediate points. Assume any element FY, draw 
FF" perpendicular to AB, then make an arc, AF", of the plan of 
the cone's base, and make A'F' = AH' ™= FF". Then Y'F' and 
YTT will be the left hand projections of the two elements project- 
ed in FY. Then project/ at /'and h' on these elements, amdg'a'h'e' 
will be the visible part of the intersection. Its right hand projec- 
tion is c/G, where / and G are, each, the projection of two points 
on opposite sides of the cone. 



so 



ELEMENTARY INTERSECTIONS. 



Remark.— By developing the cone and the cylinder, patterns 
could be made for a conical pipe entering a cylindrical one. 

64 (4). Observing that, in every case, the auxiliary planes are 
made to cut the given curved surfaces in the simplest manner, that 
is, in straight lines or circles, we have the following principles. To 
cut right lines, at once, from two cylinders, as in Fig. 5, sl plane 
must be parallel to both their axes. To cut a cylinder and cone, at 
once, in the same manner, as in Fig. 7, each plane must contain the 
vertex of the cone, and be parallel to the axis of the cylinder. To 
cut elements at once from two cones, a plane must simply contain 
both vertices. Thus, in Fig. 10, all planes cutting elements, both 
from cone VV, and cone AA', will contain the line VAB, hence 
their traces on the horizontal plane will merely pass through 
B. ^ Thus the plane BD cuts from the cone, V, the elements 
Y'a' — Ya, and VV— Vc ; and from the cone, A, the elements 
A'D' — Ad, and A'd'—AD. The student can complete the 
solution, the remainder of which is very similar to the two pre- 
ceding. 

Finally, to find the intersection of a sphere and cone, PI. III. A, 
Fig. 1 1 , auxiliary planes may most conveniently be placed in two ways' 
First, horizontally. Then each will cut a circle from the sphere 
and one from the cone ; whose horizontal projections will be circles 
and whose intersections will be points of the intersection of the 
cone and sphere. Second, vertically. Then each plane must con- 
tain the axis of the cone, from which it will cut two elements. It 
will also cut the sphere in a circle, and by revolving this plane 
about the axis of the cone till parallel to the vertical plane, as in 
Prob. 17, the intersection of the circle with the revolved elements 
see Prob. 27, may be noted, and then revolved back to their true 
position. The student can readily make the construction, after due 
familiarity with preceding problems has made the apprehension of 
the present article easy. 

Prob. 31.— To find the intersection of a vertical hexagonal 
prism with a sphere, whose centre is in the axis of the prism PI 
111. A, Fig. 8. 

Let O-ABC be part of the sphere, and DGHK the prism, 
showing one face in its real size, and therefore requiring no plan 
(48) Draw dg parallel to AC, and the arc ^A/with O as a centre, 
and through .and/. This arc is the real size of the intersection 
of the middle face of the prism with the surface of the sphere. AH 
the faces, being equal, have circular tops, equal to ehf; but, being 



ELEMENTARY INTERSECTIONS. 31 

seen obliquely, they would be really elliptical in projection. It is 
ordinarily sufficient, however, to represent them by circular arcs, 
tangent to hn, the horizontal tangent at h, and containing the 
points d and e, and /and g, as shown. 

Remark. — The heavy lines here, show the part of the prism 
within the sphere, as a spherical topped bolt head. To make 
D<fcEF, draw Od at 45° with AC, to locate dg. To make the 
spherical top flatter, for the same base DG, take a larger sphere, 
and a plane above its centre for the base of the prism. 

Prob. 32. — To construct the intersection of a vertical cone with 
a vertical hexagonal prism ; both having the same axis. PI. III. 
A, Fig. 9. 

Let VAB be the cone, and CFGH, the prism, whose elevation 
can be made without a plan (48), since one face is seen in its real 
size. The semicircle on cf is evidently equal to that of the cir- 
cumscribing circle of the base of the prism, and ct is the chord of 
two thirds of it. Then half of ct, laid off on either side of O, the 
middle of CF, as at 0;z, will give np the projection of the middle 
face EDc? after turning the prism 90° about its axis. This done, 
np will be the height, above the base, of the highest point at which 
this and all the faces will cut the cone. A vertical plane, not 
through the vertex of a cone, cuts it in the curve, or " conic sec- 
tion," called a hyperbola. The vertical edges of the prism cut the 
cone at the height ¥f hence, drawing the curves, as dse, sharply 
curved as at s, and nearly straight near d and e, we shall have a 
sufficiently exact construction of the required intersection. 

Hemark. — The heavy lines represent the part of the prism within 
the cone, finished as a hexagonal head to an iron "bolt," such as is 
often seen in machinery. The horizontal top, hg, of the head, may 
be drawn by bisecting pr at g. To make Cc=ED, as is usual in 
practice, simply draw Oc at an angle of 45° with AB, to locate cf. 
By making VAB = 30° perhaps the best proportions will be found. 

64 (5). In the subsequent applications of projections in practical 
problems, the ground line is very generally omitted; since a know- 
ledge of the object represented makes it evident, on inspection, 
which are the plans, and which the elevations. 



DIVISION SECOND. 

DETAILS OF CONSTRUCTIONS IN MASONRY, WOOD, AND METAL. 



CHAPTER I. 

CONSTRUCTIONS IN MASONRY. 

§ I. — General Definitions and Principles applicable both to JBric/e 
and Stone-work. 

65. A horizontal layer of brick, or stone, is called a course. The 
seam between two courses is called a coursing-joint. The seam 
between two stones or bricks of the same course, is a vertical or 
heading-joint. The vertical joints in any course should abut against 
the solid stone or brick of the next courses above and below. This 
arrangement is called breaking joints. The particular arrangement 
of the pieces in a wall is called its bond. As far as possible, stones 
and bricks should be laid with their broadest surfaces horizontal. 
Bricks or stones, whose length is in the direction of the length of a 
wall, are called stretchers. Those whose length is in the direction 
of the thickness of a wall, are called headers. 

§ II.— Brick Work. 

66. If it is remembered that bricks used in building have, usually, 
an invariable size, 8" x 4" X 2" (the accents indicate inches), and 
that in all ordinary cases they are used whole, it will be seen that 
brick Avails can only be of certain thicknesses, while, in the use of 
stone, the wall can be made of any thickness. 

Thus, to begin with the thinnest house wall which ever occurs, 
viz. one whose thickness equals the length of a brick, or 8 inches ; 
the next size, disregarding for the present the thickness of mortar, 
would be the length of a brick added to the width of one, or equal 
to the width of three bricks, making 12 inches, a thickness employed 
in the partition walls and upper stories of first class houses, or the 



PL II 








a 




/t\b 




\i/ 




r 


S 23. 




t P 


o' P' 












I 


b' 


X^--^ 1 




o 


S" 



CONSTRUCTIONS IN MASONRY. S3 

outside walls of small houses. Then, a wall whose thickness is 
equal to the length of two bricks or the width of four, making ] 6 
inches, a thickness proper for the outside walls of the lower stories 
of first class houses ; and lastly, a wall whose thickness equals the 
length of two bricks added to the width of one ; or, equals the 
width of five bricks, or 20 inches, a thickness proper for the base- 
ment walls of first class houses, for the lower stories of few-storied, 
heavy manufactory buildings, &c. 

67. In the common bond, generally used in this country, it may 
be observed — 

a. That in heavy buildings a common rule appears to be, to have 
one row of headers in every six or eight rows of bricks or courses, 
i.e. five or seven rows of stretchers between each two successive 
rows of headers ; and, 

b. That in the 12 and 20 inch walls there may conveniently be a 
row of headers in the back of the wall, intermediate between the 
rows of headers in the face of the wall, while in the 8 inch and 16 
inch walls, the single row of headers in the former case, and the 
double row of headers in the latter, would take up the whole thick- 
ness of the wall, and there might be no intermediate rows of 
headers. 

c. The separate rows, making up the thickness of the wall in any 
one layer of stretchers, are made to break joints in a horizontal 
direction, by inserting in every second row a half brick at the end 
of the wall. 

68. Calling the preceding arrangements common bonds, let us 
next consider the bonds used in the strongest engineering works 
which are executed in brick. These are the English bond and the 
Flemish bond. 

The English Bond. — In this form of bond, every second course, 
as seen in the face of the wall, is composed wholly of headers, the 
intermediate courses being composed entirely of stretchers. Hence, 
in any practical case, we have given the thickness of the wall and 
the arrangement of the bricks in the front row of each course, and 
are required to fill out the thickness of the wall to the best advantage. 

The Flemish Bond. — In this bond, each single course consists 
of alternate headers and stretchers. The centre of a header, in 
any course, is over the centre of a stretcher in the course next 
above or below. The face of the wall being thus designed, it 
remains, as before, to fill out its thickness suitably. 

69. Example 1. To represent an Eight Inch Wall in Eng- 
lish Bond. Let each course of stretchers consist of two rows, side 

3 



34 CONSTRUCTIONS IN MASONRY. 

by side, the bricks in which, break joints with each other hori- 
zontally. Then the joints in the courses of headers, will be distant 
half the width of a brick from the vertical joints in the adjacent, 
courses of stretchers, as may be at once seen on constructing a 
diagram. 

70. Ex. 2. To represent a Twelve Inch Wall in English 
Bond. See PI. IV., Fig. 37. In the elevation, four courses are 
shown. The upper plan represents the topmost course, and in the 
lower plan, the second course from the top is shown. The courses 
having stretchers in the face of the wall, could not be filled out by 
two additional rows of stretchers, as such an arrangement would 
cause an unbroken joint along the line, cib, throughout the whole 
height of the wall — since the courses having headers in the face, 
must be filled out with a single row of stretchers, in order to make 
a twelve inch wall, as shown in the lower plan. 

In order to allow the headers of any course to break joints with 
the stretchers of the same course, the roAV of headers may be filled 
out by a brick, and a half brick — split lengthwise — as in the upper 
plan ; or by two three-quarters of bricks, as seen in the lower 
plan. 

71. Ex. 3. To represent a Sixteen Inch Wall in English 
Bond. The simplest plan, in which the joints would overlap pro- 
perly, seems to be, to have every second course composed entirely of 
headers, breaking joints horizontally, and to have the intermediate 
courses composed of a single row of stretchers in the front and 
back, with a row of headers in the middle, which would break 
joints with the headers of the first named courses. If the stretcher 
courses were composed of nothing but stretchers, there would 
evidently be an unbroken joint in the middle of the wall extending 
through its whole height. 

72. Ex. 4. To represent an Eight Inch Wall in Flemish 
Bond. PI. IV., Fig. 38, shows an elevation of four courses, and the 
plans of two consecutive courses. The general arrangement of both 
courses is the same, only a brick, as AA', in one of them, is set six 
inches to one side of the corresponding brick, B, of the next course 
— measuring from centre to centre. 

73. Ex. 5. To represent a Twelve Inch Wall in Flemish 
Bond. PI. IV., Fig. 39, is arranged in general like the preceding 
figures, with an elevation, and two plans. One course being arranged 
as indicated by the lower plan, the next course may be made up in 
two ways, as shown in the upper plan, where the grouping shown 
at the right, obviates the use of half bricks in every second course. 



CONSTRUCTIONS IN MASONRY. 35 

There seems to be no other simple way of combining the bricks in 
this wall so as to avoid the use of half bricks, without leaving open 
spaces in some parts of the courses. 

74. Ex. 6. To represent a Sixteen Inch Wall in Flemish 
Bond. PI. IV., Fig. 40. The figure explains itself sufficiently. 
Bricks may not only be split crosswise and lengthwise, but even 
thicknesswise, or so as to give a piece 8x4x1 inches in size. 
Although, as has been remarked, whole bricks of the usual dimen- 
sions can only form walls of certain sizes, yet, by inserting frag- 
ments, of proper sizes, any length of wall, as between windows and 
doors, or width of pilasters or panels, may be, and often is, con- 
structed. By a similar artifice, and also by a skilful disposition of 
the mortar in the vertical joints, tapering structures, as tall chim- 
neys, are formed. 

§ ILL— Stone Work. 

15. The following examples will exhibit the leading varieties of 
arrangement of stones in walls. 

Example 1. Regular Bond in Dressed Stone. PL V. Fig. 41. 
Here the stones are laid in regular courses, and so that the middle 
of a stone in one course, abuts against a vertical joint in the course 
above and the course below. In the present example, those stones 
whose ends appear in the front face of the wall, seen in elevation, 
take up the w T hole thickness of the wall as seen in plan. 

The right hand end of the wall is represented as broken down in 
all the figures of this plate. Broken stone is represented by a 
smooth broken line, and the under edge of the outhanging part of 
any stone, as at n, is made heavy. 

76. Ex. 2. Irregular Rectangular Bond. PI. V., Fig. 42. In 
this example, each stone has a rectangular face in the front of the 
wall. These faces are, however, rectangles of various sizes and 
proportions, but arranged with their longest edges horizontal, and 
also so as to break joints. 

11. That horizontal line of the plan w T hich is nearest to the lower 
border of the plate, is evidently the plan of the top line of the ele- 
vation, hence all the extremities, as «', b\ &c, of vertical joints, found 
on that line, must be horizontally projected as at a and #, in the 
horizontal projection of the same line. 

18. Ex. 3. Rubble Walls. The remaining figures of PI. V., 
represent various forms of " rubble" wall. Fig. 43 represents a 
wall of broken boulders, or loose stones of all sizes, such as are 
found abundantly in New England. Since, of course, such stones 



36 CONSTRUCTIONS IN MASONEY. 

would not fit together exactly, the " chinks " between them are 
filled with small fragments, as shown in the figure. Still smaller 
irregularities in the joints, which are not thus filled, are represented 
after tinting by heavy strokes in inking. Fig. 44 represents the 
plan and elevation of a rubble wall made of slate ; hence, in the 
plan, the stones appear broad, and in the elevation, long and thin, 
with chink stones of similar shape. Fig. 45 represents a rubble 
wall, built in regular courses, which gives a pleasing effect, parti- 
cularly if the wall have cut stone corners, of equal thickness with 
the rubble courses. 

79. Execution. — The graphical construction of objects being now 
more an object of study than the mere style of execution employed, 
the graining of the stone work on PI. V. is omitted. 

Plate V. may be, 1st, pencilled; 2d, inked in fine lines; 3d, 
tinted. The rubble walls, having coarser lines for the joints, may 
better be tinted, before lining the joints in ink. 

The plans need not be tinted. 

Also, in case of the rubble walls, sudden heavy strokes may be 
made occasionally in the joints, to indicate slight irregularities in 
their thickness, as has already been mentioned. 

The right hand and lower side of any stone, not joining another 
stone on those sides, is inked heavy, in elevation, and on the plans 
as usual. The left hand lines of Figs. 43 and 44 are tangent at 
various points to a vertical straight line, walls, such as are repre- 
sented in those figures, being made vertical, at the finished end, 
by a plumb line, against which the stones rest. 

The shaded elevations on PL V., may serve as guides to the 
depth of color to be used in tinting stone work. The tint actually 
to be used, should be very light, and should consist of gray, or a 
mixture of black and white, tinged with Prussian blue, to give a 
blue gray, and carmine also if a purplish gray is desired. 

Memarks. — a. A scale may be used, or not, in making this plate. 
The number of stones shown in the width of the plans, shows that 
the walls are quite thick. 

b. Rubble walls, not of slate, are, strictly, of two kinds ; first, 
those formed of small boulders, used whole, or nearly so; and 
second, those built of broken rock, Each should show the broad- 
est surfaces in plant 



CHAPTER II. 

CONSTRUCTIONS IN WOOD. 

§ I. — General MemarJcs. 

80. Two or more beams may be framed together, so as to make 
any angle with each other, from 0° to 180°; and so that the plane 
of two united pieces may be vertical, horizontal, or oblique. 

81. To make the present graphical study of framings more fully 
rational, it may here be added, in respect to each of the positions of 
a pair of pieces, named in the last article, that they may be framed 
with reference to the resisting of forces which would act to separate 
them in the direction of any one of the three dimensions of each. 
Following out, in regular order, the two-fold classification involved 
in this and in the preceding article, let us presently proceed to notice 
several examples, some mainly by general description of their 
material construction and action, and some by a complete descrip- 
tion of their graphical construction and execution, also. 

82. Two other points may here be, however, further introduced, 
as further grounds for a systematic treatment of the present sub- 
ject. First: A pair of pieces may be immediately framed into 
each other, or they may be intermediately framed by " bolts," 
" keys," &c, or both modes may be, and often are, combined. 
Second : Two combinations of timbers which are alike in general 
appearance, may be adapted, the one to resist extension, and the 
other, compression, and may have slight corresponding diiferences 
of construction. 

83. Note. — For the benefit of those who may not have had access 
to the subject, the following brief explanation of scales, &c, is here 
inserted. 

Drawings, showing the pieces as taken apart so as to show the 
mode of union of the pieces represented, are called " Details." 

The drawings about to be described, are to be made in plan, side 
and end elevations, sections and details, or in as few of these views 
as will show clearly all parts of the object represented. 

84. In respect to the instrumental operations, these drawings are 



38 CONSTRUCTIONS IN WOOD. 

supposed to be " made to scale," from measurements of models, or 
from assumed measurements. It will, therefore, be necessary, 
before beginning the drawings, to explain the manner of sketching 
the object, and of taking and recording its measurements. 

85. In sketching the object, make the sketches in the same way 
in which they are to be drawn, i.e. in plan and elevation, and not 
in perspective, and make enough of them to contain all the mea- 
surements, i.e. to show all parts of the object. 

In measuring, take measurements of all the parts which are to be 
shown ; and not merely of individual parts alone, but such con- 
necting measurements as will locate one part with reference to 
another. 

86. The usual mode of recording the measurements, is, to indi- 
cate, by arrow heads, the extremities of the line of which the figures 
between the arrow heads show the length. 

87. For brevity, an accent (') denotes feet, and two accents (") 
denote inches. The dimensions of small rectangular pieces are 
indicated as in PI. VI., Fig. 50, and those of small circular pieces, 
as in Fig. 51. 

88. In the case of a model of an ordinary house framing, such as 
it is useful to have in the drawing room, and in which the sill is 
represented by a piece whose section is about 2J- inches by 3 inches, 
a scale of one inch to six inches is convenient. Let us then describe 
this scale, which may also be called a scale of two inches to the 
foot. 

The same scale may also be expressed as a scale of one foot to 
two inches, meaning that one foot on the object is represented by 
two inches on the drawing ; also, as a scale of £, thus, a foot being 
equal to twelve inches, 12 inches on the object is represented by 
two inches on the drawing ; therefore, one inch on the drawing 
represents six inches on the object, or, each line of the drawing is 
} of the same line, as seen upon the object ; each line, for we know 
from Geometry that surfaces are to each other as the squares of 
their homologous dimensions, so that if the length of the lilies of 
the drawing is one-sixth of the length of the same lines on the object, 
the area of the drawing would be one thirty-sixth of the area of the 
object, but the scale always refers to the relative lengths of the 
lilies only. 

89. In constructing the scale above mentioned, upon the stretched 
drawing paper, 

1st. Set off upon a fine straight pencil line, two inches, say three 
times, making four points of division. 



CONSTRUCTIONS IN WOOD. 39 

2d. Number the left hand one of these points, 12, the next, 0, 
the next, 1, the next, 2, &c, for additional points. 

3d. Since each of these spaces represents a foot, if any one of 
them, as the left hand one, be divided into twelve equal parts, 
those parts will be representative inches. Let the left hand space, 
from (12) to (0) be thus divided, by fine vertical dashes, into twelve 
equal parts, making the three, six, and nine inch marks longer, so 
as to catch the eye, when using the scale. 

4th. As some of the dimensions of the object to be drawn are 
measured to quarter inches, divide the first and sixth of the inches, 
already found, into quarters ; dividing two of them, so that each 
may be a check upon the other, and so that there need be no con 
tinual use of one of them, so as to wear out the scale. 

5th. When complete, the scale may be inked ; the length of it in 
fine parallel lines about ^V of an inch apart. 

90. It is now to be remarked that these spaces are always to be 
called by the names of the dimensions they represent, and not 
according to their actual sizes, i. e. the space from 1 to 2 repre- 
sents a foot upon the object, and is called a foot; so each twelfth 
of the foot from 12 to is called an inch, since it represents an 
inch on the object; and so of the quarter inches. 

91. Next, is to be noticed the directions in which the feet and 
inches are to be estimated. 

The feet are estimated from the zero point towards the right, 
and the inches from the same point towards the left. 

Thus, to take off 2' — 5" from the scale, place one leg of the 
dividers at 2, and extend the other to the fifth inch mark beyond 
0, to the left ; or, if the scale were constructed on the edge of a 
piece of card-board, the scale being laid upon the paper, and with 
its graduated edge against the indefinite straight line on which the 
given measurement is to be laid off, place the 2' or the 5" mark, at 
that point on the line, from which the measurement is to be laid 
off, according as the given distance is to be to the left or right 
of the given point, and then with a needle point mark the 5" point 
or the 2' point, respectively, which will, with the given point, 
include the required distance. 

92. Other scales, constructed and divided as above described, 
only smaller, are found on the ivory scale, marked 30, &c, mean- 
ing 30 feet to the inch when the tenths at the left are taken as 
feet ; and meaning three feet to the inch when the larger spaces 
— three of which make an inch — are called feet, and the twelfths 
of the left hand space, inches. Intermediate scales are marked 



40 CONSTRUCTIONS IN WOOD. 

35, etc. Thus, on the scale marked 45, four and a half of the 
larger spaces make one inch, and the scale is therefore one of four 
and a half feet to one inch, when these spaces represent feet ; 
and of forty-five feet to one inch, when the tenths represent feet. 
In like manner the other scales may be explained. 

So, on the other side of the ivory, are found scales marked -§-, 
&c, meaning scales of -| inch to one foot, or ten feet, according as 
the whole left hand space, or its tenth, is assumed as representing 
one foot. Note that -f of an inch to a foot is -§■ of a foot to the 
inch, f of an inch to ten feet, is 16 feet to an inch, &c. 

94. Of the immense superiority of drawing by these scales, over 
drawing without them, it is needless to say much : without them, 
we should have to go through a mental calculation to find the 
length of every line of the drawing. Thus, for the piece which is 
two and a half inches high, and drawn to a scale of two inches to 

a foot, we should say — 2£ inches =y| of a foot=r ¥ 5 T °f a ^ 00t * 
One foot on the object— two inches on the drawing, then ^\ of a 
foot on the object—— of 2 inches =-^ i =^ I of an inch, and ¥ 5 T of a 
foot (=2-§- inches) rz:^- of 2 inches =: T \ of an inch. 

A similar tedious calculation would have to be gone through 
with for every dimension of the object, while, by the use of scales, 
like that already described, we take off the same number of the feet 
and inches of the scale, that there are of real feet and inches in any 
given line of the object. 

§ II. — Pairs of Timbers whose axes make angles of 0° icith each 

other. 

95. Example 1. A Compound Beam "bolted, PI. II., Fig. 46. 
Mechanical Construction. 

The figure represents one beam as laid on top of another. Thus 
situated, the upper one may be slid upon the lower one in the 
direction of two of its three dimensions ; or it may rotate about 
any one of its three dimensions as an axis. A single bolt, passing 
through both beams, as shown in the figure, will prevent all of 
these movements except rotation about the bolt as an axis. Two 
or more bolts will prevent this latter, and consequently, all move- 
ment of either of the beams upon the other. A bolt, it may be 
necessary to say, is a rod of iron whose length is a little greater 
than the aggregate thickness of the pieces which it fastens toge- 
ther. It is provided at one end with a solid head, and at the other, 
with a few screw threads on which turns a " nut,'' for the purpose 



CONSTRUCTIONS IN WOOD. 4] 

of gradually compressing together the pieces through which the 
bolt passes. 

96. Graphical Construction. Assuming for simplicity's sake in 
this and in most of these examples, that the timbers are a foot square* 
and having the given scale ; the diagrams will generally explain 
themselves sufficiently. The scales are expressed fractionally, adja- 
cent to the numbers of the diagrams. The nut only is shown in the 
plan of this figure. 

It is an error to suppose that the nuts and other small parts can 
be carelessly drawn, as by hand, without injury to the drawing, 
since these parts easily catch the eye, and if distorted, or roughly 
drawn, appear very badly. 

The method is, therefore, here fully given for drawing a nut 
accurately. Take any point in the centre line, ab, of the bolt, pro- 
duced, and through it draw any two lines, cd and en, at right-angles 
to each other. From the centre, lay off each way on each line, half 
the length of each side of the nut, say f of an inch. 

Then, through the points so found, draw lines parallel to the 
centre lines cd and en, and they will form a square plan of a nut 
\y on each side. 

In making this construction, the distances should be set off very 
carefully, and the sides of the nut ruled, in very fine lines, and 
exactly through the points located. From the plan, the elevation 
is found as in PI. II., Fig. 21. 

Note. — The Teacher need not make just the same selection of 
examples, to be drawn by classes, that is given in this "Divi- 
sion ;" nor in the present collection of exercises ; in fact, wher- 
ever practicable, it would be well to take examples from real 
structures, using the specimens here given, as a guide in making 
the selections. 

97. Ex. 2. A Compound Beam, notched and bolted. PI. 
VI., Fig. 47. Mechanical Construction. The beams represented in 
this figure, are indented together by being alternately notched; the 
portions cut out of either beam being a foot apart, a foot in length, 
and two inches deep. When merely laid, one upon another, they 
will offer resistance only to being separated longitudinally, and to 
horizontal rotation. 

The addition of a bolt renders the " compound beam," thus 
formed, capable of resisting forces tending to separate it in all 
ways. 

Thin pieces are represented, in this figure, between the bolt-head 
and nut. and the wood. These are circular, having a rounded 



42 CONSTRUCTIONS IN WOOD. 

edge, and a circular aperture in the middle through which the bolt 
passes. They are called " washers" and their use is, to distribute 
the pressure of the nut or bolt-head over a larger surface, so as not 
to indent the wood, and so as to prevent a gouging of the wood 
in tightening the nut, which gouging would facilitate the decay 
of the wood, and consequently, the loosening of the nut. 

98. Graphical Construction. — The beams being understood to be 
originally one foot square, the compound beam will be 22 inches 
deep ; hence draw the upper and lower edges 22 inches apart, and 
from each of them, set off, on a vertical line, 10 inches. Through 
the points, a and #, so found, draw very faint horizontal lines, and 
on either of them, lay off any number of spaces ; each, one foot in 
length. Through the points, as c, thus located, draw transverse 
lines between the faint lines, and then, to prevent mistakes in 
inking, make slightly heavier the notched line which forms the real 
joint between the timbers. 

The use of the scale of J T continues till a new one is mentioned. 

The following empirical rales will answer for determining the 
sizes of nuts and washers on assumed sketches like those of PI. VI., 
so as to secure a good appearance to the diagram. The side of the 
nut may be double the diameter of the bolt, and the greater dia- 
meter of the washer may be equal to the diagonal of the nut, plus 
twice the thickness of the washer itself. 

Execution. — This is manifest in this case, and in most of the fol- 
lowing examples, from an inspection of the figures. 

99. Ex. 3. A Compound Beam, keyed. PI. VI., Fig. 48. 
Mechanical Construction. The defect in the last construction is, 
that the bearing surfaces opposed to separation in the direction 
of the length of the beam, present only the ends of the grain to 
each other. These surfaces are therefore liable to be readily 
abraded or made spongy by the tendency to an interlacing action 
of the fibres. Hence it is better to adopt the construction given 
in PI. VI., Fig. 48, where the " Jceys," as K, are supposed to be 
of hard wood, whose grain runs in the direction of the width of the 
beam. In this case, the bolts are passed through the keys, to pre- 
vent them from slipping out, though less boring would be required 
if they were placed midway between the keys. In this example, 
the strength of the beam is greatly increased with but a very 
small increase of material, as is proved in mechanics and confirmed 
by experiment. 

100. Graphical Construction. — This example differs from the last 
bo slightly as to render a particular explanation unnecessary. The 



CONSTRUCTIONS IN WOOD. 43 

keys are 12 inches in height, and 6 inches in width, and are 18 
inches apart from centre to centre. They are most accurately 
located by their vertical centre lines, as AA'. If located thus, and 
from the horizontal centre line BB', they can be completely drawn 
before drawing ee' and nn'. The latter lines, being then pencilled, 
only between the keys, mistakes in inking will be avoided. 

Execution. — The keys present the end of their grain to view, 
hence are inked in diagonal shade lines, which, in order to render 
the illuminated edges of the keys more distinct, might terminate, 
uniformly, at a short distance from the upper and left hand edges. 

By shading only that portion of the right hand edge of each key, 
which is between the timbers, it is shown that the keys do not 
project beyond the front faces of the timbers. 

101. Ex. 4. A Compound Beam, scarfed. PL VI., Fig. 49 
Mechanical Construction. This specimen shows the usie of a series* 
of shallow notches in giving one beam a firm hold, so to speak, 
upon another ; as one deep notch, having a bearing surface equal to 
that of the four shown in the figure, would so far cut away the 
lower beam as to render it nearly useless. 

102. Graphical Construction. — The notches, one foot long, and 
tico inches deep, are laid down in a manner similar to that described 
under Ex. 2. 

103. Execution. — The keys, since they present the end of the grain 
to view, are shaded as in the last figure. Heavy lines on their right 
hand and lower edges would indicate that they projected beyond 
the beam. 

Remark. — When the surfaces of two or more timbers lie in the 
same plane, as in many of these examples, they are said to be 
"flush"' with each other. 

§ III. — Combinations of Timbers, tohose axes make angles of 90° 
with each other. 

104. The usual way of fastening timbers thus situated, is by 
means of a projecting piece on one of them, called a " tenon," 
which is inserted into a corresponding cavity in the other, called a 
" mortise.'''' The tenon may have three, two, or one of its sides 
flush with the sides of the timber to which it belongs ; while the 
mortise may extend entirely, or only in part, through the timber 
in which it is made, and may be enclosed by that timber on three 
or on all sides. [See the examples wdnch follow, in which some of 
these cases are represented, and from which the rest can be under- 
stood.] 



44 CONSTRUCTION'S IN WOOD. 

When the mortise is surrounded on three or on two sides, par- 
ticularly in the latter case, the framed pieces are said to be 
"haloed" together, more especially in case they are of equal thick- 
ness, and have half the thickness of each cut away, as at PL VI., 
Fig. 52. 

105. Example 1. Two examples of a Floor Joist and Sill. 
(From a Model.) PI. VI., Fig. 53. 3Iechanical Construction. 
A — A' is one sill, B — B' another. CO' is a floor timber framed into 
both of them. At the left hand end, it is merely " dropped in," with 
a tenon ; at the right hand end, it is framed in, with a tenon and 
" tusk," e. At the right end, therefore, it cannot be lifted out, 
but must be drawn out of the mortise. The tusk, e, gives as great 
a thickness to be broken off, at the insertion into the sill, and as 
much horizontal bearing surface, as if it extended to the full depth 
of the tenon, t, above it, while less of the sill is cut away. Thus, 
labor and the strength of the sill, are saved. 

106. Graphical Construction. — 1st. Draw ah. 2d. On ah con- 
struct the elevation of the sills, each 2^- inches by 3 inches. 3d. 
Make the two fragments of floor timber with their upper surfaces 
flush with the tops of the sills, and 2 inches deep. 4th. The mor- 
tise in A', is f of an inch in length, by 1 inch in vertical depth. 
hth. Divide cd into four equal parts, of which the tenon and tusk 
occupy the second and third. The tenon, £, is f of an inch long, 
and the tusk, e, \ of an inch long. Let the scale of } be used. 

107. Execution. — The sills, appearing as sections in elevation, are 
shaded. In all figures like this, dotted lines of construction should 
be freely used to assist in " reading the drawing," i.e. in com- 
prehending, from the drawing, the construction of the thing repre- 
sented. 

108. Ex. 2. Example of a " Mortise and Tenon," and of 
"Halving." (From a Model.) PI. VI., Fig. 54. Mechanical 
Construction. In this case, the tenon, AA', extends entirely 
through the piece, CO', into which it is framed. B and C are 
halved together, by a mortise in each, whose depth equals half 
the thickness of B, as shown at B" and C", and by the dotted 
line, ab. 

Graphical Construction. — Make, 1st, the elevation, A'; '2d, the 
plan ; 3d, the details. B" is an elevation of B as seen when 
looking in the direction, BA. C" is an elevation of the left hand 
portion of CC, showing the mortise into which B is halved. The 
dimensions may be assumed, or found by a scale, as noticed below. 

109. Execution. — The invisible parts of the framing, as the halv- 



CONSTRUCTIONS IN WOOD. 45 

ing, as seen at db in elevation, are shown in dotted lines. The 
brace and the dotted lines of construction serve to show what 
separate figures are comprehended under the general number (54) 
of the diagram. The scale is }. From this the dimensions of the 
pieces can be found on a scale. 

110. Ex. 3. A Mortise and Tenon as seen in two sills 
and a post. Use of broken planes of section. (From a 
Model.) PI. VI., Fig. 55. 

Mechanical Construction. — The sills, being liable to be drawn 
apart, are pinned at a. The post, BB', is kept in its mortise, bb'\ by 
its own weight ; m is the mortise in which a vertical wall joist 
rests. It is shown again in section near m' . 

111. Graphical Construction. — The plan, two elevations, and a 
broken section, show all parts fully. 

The assemblage is supposed to be cut, as shown in the plan by 
the broken line AA'A"A"', and is shown, thus cut, in the shaded 
figure, A'A'A'"m'. The scale, which is the same as in Fig. 53, 
indicates the measurements. At B", is the side elevation of the 
model as seen in looking in the direction AA. 

In Fig. 55 a, A's obviously equals A"s, as seen in the plan. 

112. Execution. — In the shaded elevation, Fig. 55a, the cross-sec- 
tion, AA"', is lined as usual. The longitudinal sections are shaded 
by longitudinal shade lines. The plan of the broken upper end of 
the post, B, is filled with arrow heads, as a specimen of a way, 
sometimes convenient, of showing an end view of a broken end. 

Sometimes, though it renders the execution more tedious, narrow 
blank spaces are left on shaded ends, opposite to the heavy lines, 
so as to indicate more plainly the situation of the illuminated edges 
(100). The shading to the left of A', Fig. 55a, should be placed 
so as to distinguish its surface from that to the right of A'. 

113. Ex. 4. A Mortise and Tenon, as seen in timbers so 
framed that the axis of one shall, when produced, be a 
diagonal diameter of the other. PI. VI., Fig. 56. Mechani- 
cal Construction. — In this case the end of the inserted timber is not 
square, and in the receiving timber there is, besides the mortise, a 
tetraedron cut out of the body of that timber. 

114. Graphical Construction. — D is the plan, D' the side eleva- 
tion, and D" the end elevation of the piece bearing the tenon. F' 
and F are an elevation and plan of the piece containing the mor- 
tise. Observe that the middle line of D, and of D', is an axis of 
symmetry, and that the right hand edges of D and D' are parallel 
to the sides of the incision in F'. 



46 CONSTRUCTIONS IN WOOD. 

§ IV. — Miscellaneous Combinations. 

115. Example 1 . Do welling. (From a Model.) PI. VI., Fig. 5 7. 
Mechanical Construction. — Dowelling is a mode of fastening by 
pins, projecting usually from an edge of one piece into correspond- 
ing cavities in another piece, as seen in the fastening of the parts 
of the head of a water tight cask. The mode of fastening, how- 
ever, rather than the relative position of the pieces, gives the name 
to this mode of union. 

The example shown in PI. VI., Fig. 57, represents the braces of 
a roof framing as do welled together with oak pins. 

116. Graphical Construction. — This figure is, as its dimensions 
indicate, drawn from a model. The scale is one-third of an inch 
to an inch. 

1st. Draw acb, with its edges making any angle with the imagi- 
nary ground line — not drawn. 

2d. At the middle of this piece, draw the pin or dowel, pp, \ of 
an inch in diameter, and projecting f of an inch on each side of the 
piece, acb. This pin hides another, supposed to be behind it. 

3d. The pieces, d and d'', are each 2-J- inches by 1 inch, and are 
shown as if just drawn off from the dowels, but in their true direc- 
tion, i.e. at right angles to acb. 

4th. The inner end of d is shown at d\ showing the two holes, 
1^- inches apart, into which the dowels fit. 

Execution. — The end view is lined as usual, leaving the dowel 
holes blank. 

117. Ex. 2. A dovetailed Mortise and Tenon. PI. VI., 
Fig. 58. Mechanical Construction. — This figure shows a species of 
joining called dovetailing. Here the mortise increases in width as 
it becomes deeper, so that pieces which are dovetailed together, 
either at right angles or endwise, cannot be pulled directly apart. 
The corners of drawers, for instance, are usually dovetailed ; and 
sometimes even stone structures, as lighthouses, which are exposed 
to furious storms, have their parts dovetailed together. 

118. Graphical Construction. — The sketches of this framing are 
arranged as two elevations. A bears the dovetail, B shows the 
length and breadth of the mortise, and B" its depth. A and B 
belong to the same elevation. 

Execution. — In this case a method is given, of representing a 
hidden cut surface, viz. by dotted shade lines, as seen in the hidden 
faces of the mortise in B". 

119. Leaving now the examples of pieces framed together at 
right angles let us consider : — 



CONSTRUCTIONS IN WOOD. 4^ 



§ V. —Pairs of Timbers which are framed together obliquely to 

each other. 

Example 1. A Chord and Principal. (From a Model.) PI. 
VII., Fig. 59. Jfechanical Construction. — The oblique piece 
(" principal ") is, as the two elevations together show, of equal 
width with the horizontal piece ("chord," or "tie beam"), and is 
framed into it so as to prevent sliding sidewise or lengthwise. 

Neither can it be lifted out, on account of the bolt which is made 
to pass perpendicularly to the joint, ac, and is "chipped up" at pp, 
so as to give a flat bearing, parallel to ac, for the nut and bolt-head. 

120. Graphical Construction. — 1st. Drawee/ 2d. Lay off de 
= 13 inches; 3c?. Make e'ea=30° ; 4th. At any point, e\ draw a 
perpendicular to ee', and lay off upon it 9 inches — the perpendicular 
width of e'ea; 5th. Make ec = 4 inches and perpendicular to e'e ; 
bisect it and complete the outlines of the tenons, and the shoulder 
aec' ; QtJi. To draw the nut accurately, proceed as in PI. VI., Fig. 
46-47, placing the centre of the auxiliary projection of the nut in 
the axis of the bolt produced, &c. (46) (96). 

Execution. — b represents the bolt hole, the bolt being shown 
only on one elevation. 

121. Ex. 2. A Brace, as seen in the angle between a 
"post" and "girth." (From a Model.) PI. VII., Fig. 60. 
Mechanical Construction. — PP' is the post, GG' is the girth, and 
B'B" is the brace, having a truncated tenon at each end, which rests 
in a mortise. When the brace is quite small, it has a shoulder on 
one side only of the tenon, as if B'B" were sawed lengthwise on 
a line, oo' . 

122. Graphical Construction. — To show a tenon of the brace 
clearly, the girth and brace together are represented as being 
drawn out of the post. 1st. Draw the post. 2c?. Half an inch 
below the top of the post, draw the girth 2^ inches deep. 3d. From 
a, lay off ab and a each 4 inches, and draw the brace 1 inch wide. 
4th. Make cd equal .to the adjacent mortise; viz. 1^ inches; make 
de—l inch, and erect the perpendicular at e till it meets be, etc. 
The dotted projecting lines show the construction of B" and of the 
plan. At e" is the vertical end of the tenon e. On each side of 
c ff , are the vertical surfaces, shown also at cd. 

123. Ex. 3. A Brace, with shoulders mortised into the 
post. PI. VII., Fig. 51. This is the strongest way of framing a 
brace. For the rest, the figure explains itself. Observe, however, 
that while in Fig. 59, the head of the tenon and shoulder is perpendi- 



48 CONSTRUCTIONS IN WOOD. 

cular to the oblique piece, here, where that piece is framed into a 
vertical post, its head nu is perpendicular to the axis of the post. 
In Fig. 61, moreover, the auxiliary plane on which the brace 
alone is projected, is parallel to the length of the brace, as is 
shown by the situation of B, the auxiliary projection of the brace," 
and by the direction of the projecting lines, as rr\ P' is the 
elevation of P, as seen in the direction nn', and with the brace 
removed. 

124. Ex. 4. A "Shoar." PI. VII., Fig. 62. Mechanical Con- 
struction. — A " shoar " is a large timber used to prop up earth or 
buildings, by being framed obliquely into a horizontal beam 
and a stout vertical post. It is usually of temporary use, during 
the construction of permanent works ; and as respects its action, 
it resists compression in the direction of its length. To* give a 
large bearing surface without cutting too far into the vertical tim- 
ber, it often has two shoulders. The surface at ab is made verti- 
cal, for then the fibres of the post are unbroken except at cb, while 
if the upper shoulder were shaped as at adb, the fibres of the trian- 
gular portion dbc would be short, and less able to resist a longitu- 
dinal force. 

125. The Graphical construction is evident from the figure, 
which is in two elevations, the left hand one showing the post 
only. 

Execution. — The vertical surface at ab may, in the left hand 
elevation, be left blank, or shaded with vertical lines as in PL VL, 
Fig. 5 5 a. 

§ VI. — Combinations of Timbers whose axes malce angles oflQ0° 
with each other. 

126. Timbers thus framed are, in general, said to be spliced. Six 
forms of splicing are shown in the following figures. 

Example l. A Halved Splicing, pinned. PL VII., Fig. 63. 
The mechanical construction is evident from the figure. When 
boards are lapped on their edges in this way, as in figure 69, they 
are said to be " rabbetted." 

127. Graphical Construction. — After drawing the lines, 12 inches 
apart, which represent the edges of the timber, drop a perpendicu- 
lar of 6 inches in length from any point as a. From its lower ex- 
tremity, draw a horizontal line 12 inches in length, and from c, drop 
the perpendicular cb, which completes the elevation. In the plan, 
the joint at a will be seen as a full line a'a\ and that at Z>, being 
hidden, is represented as a dotted line, at b" 





































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1 



CONSTRUCTIONS IN WOOD. 49 

Draw a diagonal, a'b\ divide it into four equal parts, and take 
the first and third points of division for the centres of two pins, 
having each a radius of three-fourths of an inch. 

Execution. — The position of the heavy lines on these figures is 
too obvious to need remark. 

128. Ex. 2. Tonguing and Grooving; and Mortise and 
Tenon Splicing. PI. VII., Fig. 64. Boards united at their edges 
in this way, as shown in PL VII., Fig. 70, are said to be tongued 
and grooved. 

Drawing, as before, the plan and elevation of a beam, a foot 
square, divide its depth an, at any point «, into five equal parts. 
Take the second and fourth of these parts as the width of the 
tenons, which are each a foot long. 

The joint at a is visible in the plan, the one at b is not. Let a'd 
be a diagonal line of the square a'd. Divide ad into three equal 
parts, and take the points of division as centres of inch bolts, with 
heads and nuts 2 inches square, and washers of If inches radius. To 
place the nut in any position on its axis, draw any two lines at right 
angles to each other, through each of the bolt centres, and on each, 
lay off 1 inch from those centres, and describe the nut. Project up 
those angles of the nut which are seen ; viz. the foremost ones, 
make it 1 inch thick in elevation, and its washer J an inch thick. 

In this, and all similar cases, the head of the bolt, 5, would not 
have its longer edges necessarily parallel to those of the nut. To 
give the bolt head any position on its axis, describe it in an auxiliary 
plan just below it. 

129. Ex. 3. A Scarfed Splicing, strapped. PI. VII., Fig. 65. 
Mechanical Construction. — While timbers, framed together as in 
the two preceding examples, can be directly slid apart when their 
connecting bolts are removed ; the timbers, framed as in the present 
example, cannot be thus separated longitudinally, on account of tha 
dovetailed form of the splice. Strapping makes a firm connexion, 
but consumes a great deal of the uniting material. 

130. Graphical Construction. — After drawing the outlines of the 
side elevation, make the perpendiculars at a and a\ each 8 inches 
long, and make them 18 inches apart. One inch from a, make the 
strap, ss r , 2^ inches wide, and projecting half an inch — i. e. its 
thickness — over the edges of the timber. The ear through which 
the bolt passes to bind the strap round the timber, projects two 
inches above the strap. Take the centre of the ear as the centre 
of the bolt, and on this centre describe the bolt head l£ inches 
square. 

4 



50 CONSTRUCTIONS IN WOOD. 

In the plan, the ears of the strap are at any indefinite distance 
apart, depending on the tightness of the nut, s. 

A fragment of a similar strap at the other end of the scarf, is 
shown, with its visible ends on the bottom, near a', and back, at r y 
of the beam. 

Execution. — The scarf is dotted where it disappears behind the 
strap ; and so are the hidden joint at a\ and the fragment of the 
second strap, as shown in plan. These examples may advan- 
tageously be drawn by the student on a scale of ^. Care must 
then be exercised in making the large broken ends neatly, in large 
splinters, edged with fine ones. 

131. Ex. 4. A Scarfed Splicing; notched, keyed, and 
bolted. PI. VII., Fig. 66. Mechanical Construction. — This is a 
specimen of a very secure framing. The notches e and e' prevent la- 
teral displacement, and the keys of hard wood give a better bearing 
surface than when the ends of the grain bear upon each other (99). 

132. Graphical Construction. — Making the outlines of the timber 
1 foot square, take a'b' -= 3 feet ; draw the perpendicular bb\ and con- 
nect a' and b'. From a!- lay off* 4 inches on a'b\ and divide the remainder 
of a'b' into three equal parts ; then at &', and c, draw perpendiculars, 
each 2 inches long, above a'b'. Now divide a'c into two equal 
parts, and from c and d lay off two inches towards a', and, on these 
distances, complete the squares which represent the keys at those 
points, the one at d being below a'b'. Lastly, draw, at a\ the joint 
line perpendicular to a'b' . 

In the plan, let the angles, as eao— 60°, a being projected down 
from a\ and e being in the middle of the width of the plan. 

Let the bolts, as nn\ be so placed that their axes shall bisect a'd 
and b'c ; and let the washers be 5 inches in* diameter, and the nuts 
2| inches square, by \\ inches thick. 

a', being horizontally projected at a and a% a"v is the horizontal 
projection of a'v'. Project e at e" and draw e"s\ parallel to a'v' 
till it meets v's', parallel to a'd. Then project s' at s, and draw sv, 
the horizontal projection of sV. Similarly sv" is drawn, se is the 
horizontal projection of e"s. 

a"e and vs will, of course, not be parallel, though this is indis- 
tinctly shown in so small a figure. 

Execution. — The keys are shaded. The hidden cut surfaces of 
the notches are shaded in dotted shade lines, and the hidden joints 
are dotted. 

133. Ex. 5. A Compound Beam, with one of the compo- 
nent beams " fished." PI. VII., Fig. 67. Mechanical Construe- 



CONSTRUCTIONS IN WOOD. 51 

ti on , — The mode of union called fishing, consists in uniting two 
pieces, end to end, by laying a notched piece over the joint and 
bolting it through the longer pieces. 

The figure shows this mode as applied to a compound beam, 
i. e. to a beam " built " of several pieces bolted and keyed toge- 
ther. The order of construction is as follows — taking for a scale 
| of an inch to a foot. 

134. Graphical Construction. — 1st. The outside lines of the plan 
are two feet apart, the outside pieces are each 4£ inches wide, 
and the interior ones 5j inches ; leaving four inches for the sum 
of the three equal spaces between the four beams. 

2d. Let there be a joint at a'. Lay off 3 feet, each side of d for 
the length of the "fish." 

3d. The straight side of this piece is let into the whole piece, b, 
two thirds of an inch, and into a\ 1 inch, making its thickness 3 
inches. 

4th. At a\ it is two inches thick, i. e. at a', the .timber, a', is of 
its full thickness. The fish, c, is 2 inches thick for the space of one 
foot at each side of a'. The notches at d and e are each 1 inch 
deep, dd' and ee' each are one foot, and the notches at d' and e' 
are each 1 inch deep. The remaining portions of the fish are 1 
foot long, and 3 inches wide. 

5th. Opposite to these extreme portions, are keys, 1 foot by 3 
inches, in the spaces between the other timbers, and setting an equal 
depth into each timber. 

6th. In elevation, only the timber a' is seen — 1 foot deep. Four 
bolts pass through the keys. Vb" —h feet, and b' is three inches from 
the top, and from hk\ the left hand end of the fish. n'n" — h feet, 
and n' is 3 inches above the bottom of the timber, and 9 inches from 
Ick'. The circular bolt head is one inch in diameter, and its washer 
3^- inches diameter and J an inch thick. The thickness of the bolt 
head, as seen in plan, is f inch. The nuts, nn, are l£ inches square, 
and % an inch thick, and the bolts are half an inch in diameter. 

135. The several nuts would naturally be found in various posi- 
tions on their axes. To construct them thus with accuracy, as seen 
in the plan, one auxiliary elevation, as N, is sufficient. N", and its 
centre, may be projected upon as many planes— xy— as there are 
different positions to be represented in the plan, each plane being 
supposed to be situated, in reference to N, as some nut in the plan 
is, to its elevation. Then transfer the points on xy, &c, to the 
outside of the several ra^-washers, placing the projection of the 
centre lines of the bolts in the plan, as lines of reference. 



52 CONSTRUCTIONS IN WOOD. 

Execution. — The figure explains itself in this respect. 

136. Ex. 6. A vertical Splice. PI. VIII., Fig. 68. Mecha- 
nical Construction. — This splice is formed of two prongs at oppo- 
site corners of each piece, embraced by corresponding notches in 
the other piece. Thus in the piece B', the visible prong, as seen in 
elevation, is a truncated triangular pyramid whose horizontal base 
is abc — a'c\ and whose oblique base is enc — e'n"c"; Besides the 
four prongs, two on each timber, there is a flat surface abfq — c'dq\ 
well adapted to receive a vertical pressure, since it is equal upon, 
and common to, both timbers. 

137. Graphical Construction. — To aid in understanding this 
combination, an oblique projection is given on a diagonal plane, 
parallel to PQ. 

1st. Make the plan, acfq, with the angles of the interior square 
in the middle of the sides of the outer one. 2c?. Make the distances, 
as ce = 2 inches and draw en, &c. 3c?. Make c'n' — c'n" 15 inches, 
and draw short horizontal lines, n"e\ &c, on which project e, &c, 
after which the rest is readily completed. 

138. Execution. — Observe, in the plans, to change the direction 
of the shade lines at every change in the position of the surface of 
the wood. 



CHAPTER ni. 



CONSTRUCTIONS IN METAL. 



139. Example 1. An end view of a Railroad Rail. PI. 
VII., Fig. 7 1 . Graphical Construction. — 1 st. Draw a vertical centre 
line AA', and make AA'=3f inches. 2c?. Make A'b=A'b' = 2 
inches. 3d. Make Ac = Ac' = 1 inch. 4th. Describe two quadrants, 
of which c'd is one, with a radius of half an inch. 5th. With p, 
half an inch from AA', as a centre, and pd as a radius, describe an 
arc, dr, till it meets a vertical line through e. 6th. Draw the tangent 
rs. 7th. Draw a vertical line, as pt, % an inch from AA', on each 
side of AA'. Sth. Bisect the angle rst' and note s', where the 
bisecting line meets the radius, pr, produced. 9th. With s r as a 
centre, draw the arc rt. 10th. At b and b' erect perpendiculars, 
each one fourth of an inch high. 11th. Draw quadrants, as q't, 
tangent to these perpendiculars and of one fourth of an inch radius. 
\2th. Draw the horizontal line tv. 13th. Make nt r =nv and describe 
the arc t'v. 14th. Repeat these operations on the other side of the 
centre line, AA'. 

Execution. — Let the construction be fully shown on one side of 
the centre line. 

140. Ex. 2. An end elevation of a Compound Rail. PI. VII., 
Fig. 72. Mechanical Construction. — The compound rail, is a rail 
formed in two parts, which are placed side by side so as to break 
joints, and then riveted together. As one half of the rail is whole, 
at the points where a joint occurs on the other half, the noise and 
jar, observable in riding on tracks built in the ordinary manner, are 
both obviated ; also " chairs," the metal supports which receive the 
ends of the ordinary rails, may be dispensed with, in case of the 
use of the compound rail. 

In laying a compound rail on a curve, the holes, through which 
the bolts pass, may be drawn past one another by the bending 
of the rails. To allow for this, these holes are " slotted, 1 ' as it is 
termed, i. e. made longer in the direction of the length of the rail. 

141. Graphical Construction. — 1st. Make ty^=4 inches. 2d. 
Bisect ty at u, and erect a perpendicular, ua, of 3\ inches. 3d. 



54 CONSTRUCTIONS IN METAL. 

Make, successively, ur=% an inch; from r to cb—2 inches; from u 
to nh—^ of an inch ; and to ge = 2% inches. 4th. For the several 
widths of the interior parts, make bc~^ of an inch, and g and e 
each § of an inch, from ua ; nh=ge, and ro=§ of an inch. 5th. 
To locate the outlines of the rail, make ms, the flat top, called the 
tread of the rail, =2 inches, half an inch below this, make the width, 
/#, 3 inches; and make the part through which the rivet passes, 1| 
inches thick, and rounded into the lower flange which is J of an 
inch thick. 

The rivet has its axis If inches from ty. Its original head, q, is 
conical, with bases of— -say 1 inch, and f of an inch, diameter ; and 
is half an inch thick. The other head, jt>, is made at pleasure, being 
roughly hammered down while the rivet is hot, during the process 
of track-laying. A thin washer is shown under this head. 

142. Ex. 3. A "Cage Valve," from a Locomotive Pump. 
PI. VII., Fig. 73-74. — Mechanical Construction. — This valve is 
made in three pieces, viz. the valve proper, Fig. 74 ; the cage con- 
taining it, Abb' ; and the flange bb'c ; whose cylindrical aperture — 
shown in dotted lines — being smaller than the valve, confines it. 
The valve is a cup, solid at the bottom, and makes a water tight 
joint with the upper surface of the flange, inside of the cage. The 
whole is inclosed in a chamber communicating with the pump 
barrel, and witsb the tender, or the boiler, according as we suppose 
it to be the inlet or outlet valve of the pump. This chamber makes 
a water tight joint with the circumference of the flange bb! 

Suppose the valve to be the latter of the two just named. The 
" plunger " of the pump being forced in, the water shuts the inlet 
valve, and raises the outlet valve, and escapes between the bars of 
the cage into the chamber, and from that, by a pipe, into the boiler. 

143. Graphical Construction. — Construct the plan first, in which 
the bars appear as six equal and equidistant wings, which are then 
projected into the elevation. The diameters of the circles seen in 
the cage are, in order, from the centre, 1£, 2 J, 2)f, and 4 inches. 
The thickness of the valve is T \ of an inch, and its outside height If 
inches, and the outside diameter 2 T \ inches. The diameter of the 
aperture in the flange is If inches, its length f of an inch, and the 
height of the whole 3 T 3 g inches. 

144. Execution. — Observe carefully the position of the heavy 
lines on the elevation, which conform to the rational rule of being 
the projections of those edges of the object itself, which divide its 
illuminated from its dark surfaces. The section, Fig. 74, being of 
metal, is appropriately shaded more finely than sections of wood. 



CONSTRUCTIONS IN METAL. 55 

145. Ex. 4. An oolique elevation of a Bolt Head. PI. 
VII., Fig. 75. Let PQ be the intersection of two vertical planes, 
at right angles with each other ; and let RS be the intersection, 
with the vertical plane of the paper to the left of PQ, of a plane 
which, in space, is parallel to the square top of the bolt head. On 
such a plane, a plan view of the bolt head may be made, showing 
two of its dimensions in their real size; and on the plane above RS, 
the thickness of the bolt head, and diameter of the bolt, are shown 
in their real size. 

Below RS, construct the plan of the bolt head, with its sides 
making any angle with the ground line RS. Project its corners in 
perpendiculars to RS, giving the left hand elevation, w T hose thick- 
ness is assumed. 

146. The fact that the projecting lines of a point, form, in the 
drawings, a perpendicular to the ground line, is but a special case 
of a more general truth, which may be thus stated. — When an 
object in space is projected upon any two planes which are at right 
angles to each other, the projecting lines of any point of that object 
form a line, in the drawing, perpendicular to the intersection of the 
two planes. 

147. To apply the foregoing principle to the present problem ; 
it appears that each point, as a", of the right hand elevation, will 
be in a line, a'a", perpendicular to PQ, the intersection of the two 
vertical planes of projection. 

Remembering that PQ is the intersection of a vertical plane — 
perpendicular to the plane of the paper — with the vertical plane 
of the paper, and observing that the figure represents this plane as 
being revolved around PQ towards the left, and into the plane of 
the paper, and observing the arrow, which indicates the direction 
in which the bolt head is viewed, it appears that the revolved ver- 
tical plane, has been transferred from a position at the left of the 
plan acne, to the position, PQ, and that the centre line tic", must 
appear as far from PQ as it is in front of the plane of the paj)er — 
i. e. e"u" =eu, showing also, that ase — e' is in the plane of the paper, 
its projection at e" must be in PQ, the intersection of the two ver- 
tical planes. 

Similarly, the other corners of the nut, as c", n", <fcc., are laid off 
either from the centre line tu", or from PQ. Thus v"c''=vc, or 
v'"c" — so. The diameter of the bolt is equal in both elevations. 

148. Other supposed positions of the auxiliary plane PQ may be 
assumed by the student, and the corresponding construction worked 
out. Thus, the primitive position of PQ may be at the right of the 



56 CONSTRUCTIONS IN METAL. 

bolt head, and that may be viewed in the opposite direction from 
that indicated by the arrow. 

149. Ex. 5. A " Step" for the support of an oblique tim- 
ber. PL VII., Fig. 76. Mechanical Construction. — It will be 
frequently observed, in the framings of bridges, that there are 
certain timbers whose edges have an oblique, direction in a vertical 
plane, while at their ends they abut against horizontal timbers, not 
directly, for that would cause them to be cut off obliquely, but 
through the medium of a prismatic block of wood or iron, so 
shaped that one of its faces, as ab — n'b', Fig. 76, rests on the hori- 
zontal timber, while another, as ac — e"d"c", is perpendicular to the 
oblique timber. 

To secure lightness with strength, the step is hollow underneath, 
and strengthened by ribs, rr. The holes, h'h", allow the passage 
of iron rods, used in binding together the parts of the bridge. These 
holes are here prolonged, as at h, forming tubes, which extend partly 
or wholly through the horizontal timber on which the step rests, in 
order to hold the step steadily in its place. 

150. RemarJc. When the oblique timber, as T, Fig. 76 (a), sets 
into, rather than upon, its iron support S, so that the dotted lines, 
ab and cb, represent the ends of the timber, the support, S, is called 
a shoe. 

151. Graphical Construction. — In the plate, abc is the elevation, 
and according to the usual arrangement would be placed above the 
plan, e /r n"c", of the top of the step, a'b'e is the plan of the under 
side of the step, showing the ribs, &c. A line through nm is a 
centre line for this plan and for the elevation. A line through the 
middle point, r, of win, is another centre line for the plan of the 
bottom of the step. Having chosen a scale, the position of the 
centre lines, and the arrangement of the figures, the details of the 
construction may be left to the student. 

152. Ex. 6. A metallic steam tight " Packing," for the 
"stuffing boxes" of piston rods. PI. VII., Fig. 77. 

Mechanical Construction. — Attached to that end, b, of Fig. 77 (a), 
of a steam cylinder, for instance, at which the piston rod, p, enters 
it, is a cylindrical projection or " neck," n, having at its outer end 
a flange, ff through which two or more bolts pass. At its inner 
end, at p, this neck fits the piston rod quite close for a short space. 
The internal diameter of the remaining portion of the neck is suffi- 
cient to receive a ring, rr, which fits the piston rod, and has on its 
outer edge a flange, t, by which it is fastened to the flange, ff on 
the neck of the cylinder by screw bolts. The remaining hollow 



pl.v: 



lM-^H^d-^^>^c/ 




CONSTRUCTIONS IN METAL. 57 

space, s, between the ring or " gland," tr, and the inner end of the 
neck, is usually filled with some elastic substance, as picked hemp, 
which, as held in place by the gland, tr, makes a steam tight joint; 
which, altogether, is called the "stuffing box." 

153. The objection to this kind of packing is, that it requires so 
frequent renewals, that much time is consumed, for instance in rail- 
road repair shops, in the preparation and adjustment of the packing. 
To obviate this loss of time, and perhaps because it seems more 
neat and trim to have all parts of an engine metallic, this metallic 
packing, Fig. 77, was invented. ABC — A'B' is a cast iron ring, 
cylindrical on the outside, and having inlaid, in its circumference, 
bands, tt\ of soft metal, so that it may be squeezed perfectly tight 
into the neck of the cylinder. The inner surface of this ring is coni- 
cal, and contains the packing of block tin. This packing, as a whole, 
is also a ring, whose exterior is conical, and fits the inner side of the 
iron ring, and whose interior, /Vfcc, is cylindrical, fitting the piston 
rod closely. 

154. For adjustment, this tin packing is cut horizontally into 
three rings, and each partial ring is then cut vertically, as shown in 
the figure, into two equal segments, abed — a'b'c'd', is one segment 
of the upper ring; efgh — e'f'f"g'ti,\s one segment of the middle ring, 
and ijkl — i'j'Tc'n'l', is one segment of the lower ring. The segments 
of each ring, it therefore appears, break joints with each of the other 
rings. Three of the segments, one in each partial ring, are loose, 
while the other three are dowelled by small iron pins, parallel to 

ii the axis of the whole packing. 

I 155. Operation. — Suppose the interior,/*^, of the packing to be 
of less diameter than the piston rod, which it is to surround. By 
drawing it partly out of its conical iron case, the segments forming 
each ring can be slightly separated, making spaces at ab, &c, which 
will increase the internal diameter, so as to receive the piston rod. 
"When in this position, let the gland, tr, be brought to bear on the 
packing, and it will be firmly held in place ; then, as the packing 
gradually wears away, the gland, by being pressed further into the 
neck, will press the packing further into its conical seat, which 
will close up the segments round the piston rod. 

156. Graphical Construction. — Let the drawing be on a scale 
of one half the original size of the packing for a locomotive valve 
chest. C is the centre for the various circles of the plan, and IXD', 
projected up from C, is a centre line for portions of the elevation. 
Cf=^j of an inch ; Ce=J-| of an inch, and CA=lJ inch. A'n=2 
inches, and AV= T 9 -g- of an inch. The iron case being constructed 



58 CONSTRUCTIONS IN METAL. 

from these measurements, the rings must be located so that fp\ for 
instance, shall be = , 7 g of an inch ; and then let the thickness, f'f\ 
of each segment be \\ of an inch. These dimensions and the con- 
sequent arrangements of the rings will give spaces between the seg- 
ments, as at ab, of } of an inch ; though in fact, as this space is 
Variable, there is no necessity for a precise measurement for it. In 
the plan, there are shown one segment, and a fragment, cibef, of 
another, in the upper ring ; one segment and two fragments of the 
middle ring, and both segments of the lower ring, with the whole 
of the iron case. 

157. Execution. — The section lines in the elevation indicate clearly 
the situation of the three segments, abed, efgh, and ijkl, there shown. 
The dark bands on the case at t and t\ indicate the inlaid bands ol 
soft metal already described. 



DIVISION THIRD. 

RUDIMENTS OF SHADES AND SHADOWS. 



CHAPTER I. 

SHADOWS. 

§ I. — Observed Facts and Practical Rules. 

158. The lines of light, called rays, which emanate from a lumi- 
nous body, proceed in straight lines. This is proved by the fact that 
the shadow of a straight line, as the corner of a house, is a straight 
line, when cast upon a plane having any position relative to the 
line. 

159. If the student can see from his window a house, part of 
whose body projects beyond the rest, and a chimney upon a flat 
roof, he can learn, by simple inspection, many of those natural facts 
respecting shadows which follow directly from the preceding article, 
and are at the foundation of the solution of most of the simple pro- 
blems which occur in roof or bridge drawing, where the longer 
shadows are most frequently cast by straight lines on plane 
surfaces. 

160. It will be thus observed that the shadow of a vertical edge, 
ab, PI. VIIL, Fig. 78, of the body of the house will be a vertical 
line, a'b\ on the front wall of the wing behind it; that the shadow 
of a horizontal line, as be — the arm for a swinging sign — which is 
parallel to the wing wall, will be a horizontal line, b'e\ parallel to 
be j that the shadow of a horizontal line, be, which is perpendicular 
to the wing wall, will have an oblique shadow, cb\ commencing at 
c, where the line pierces the wing wall, and ending at b\ where a 
ray of light through b pierces the wing wall ; and finally that the 
shadow of a point, b, is at b' where the ray bb\ through that point, 
pierces the surface receiving the shadow. 

161. Passing now to PL VIII., Fig. 79, which represents a chim- 



60 SHADOWS 

ney upon aflat roof, we observe that the shadows of be and cd — lines 
parallel to the roof — are b'c' and c'd', lines equal to, and parallel to, 
the lines be and cd ; and that the shadows of ab and ed are ab f and 
ed' — similar to the shadowed' in Fig. 78 — i. e. commencing at a and 
e, where the lines casting them meet the roof, and ending at b' and 
d', where rays through b and d meet the roof. 

162. By considering the facts expressed in the two preceding 
articles, we may arrive at three simple general principles; viz. 1st. 
The shadow of a point on any surface, is where a ray uf light drawn 
through that point meets that surface. 2nd. The shadow of a 
straight line on a plane, parallel to it, is a parallel straight line ; and 
2>rd. The shadow of aline which is perpendicular to a plane, is in the 
direction of the projection of a ray of light on that plane ; for in 
Fig. 78, be, being perpendicular to the wing wall, is a projecting 
line of the ray bb' upon that wall, and b' , where the ray meet-s the 
wall, is another point of its projection upon the wall; hence cb' 
is the projection of the ray bb' on the wing wall. So in PI. VIII., 
Fig. 79, the planes of the triangles abb' and edd' are perpen- 
dicular to the top of the roof; hence the shadow lines ab' and 
ed' are projections of the rays bb' and dd', which limit those 
shadows. 

163. By further examination, the following truths become appa- 
rent ; 1st. The shadow of a circle on a plane parallel to itself, is an 
equal circle. 2nd. The point, as at a', PI. VIII., Fig. 78, where a 
shadow — aa' — on one surface intersects a second surface, is a point 
of the shadow of the same line — ab, on the second surface. 

164. Passing now to the graphical solution of problems of sha- 
dows, there appear certain necessary conditions to be known, and 
certain practical rules to be observed. 

The necessary conditions for solving a problem in shadows are 
three: — 1st, the position of the body casting the shadow ; 2nd, the 
position of the surface receiving the shadow ; and 3rd, a given 
direction of the light. 

In respect to the above conditions, it may be remarked: — 
1st, that the body casting the shadow is given by its projections, 
made according to the methods of Division I. ; 2nd, that the sur- 
faces, one or more, which receive the shadow may be the planes of 
projection themselves, or other surfaces represented by their projec- 
tions on those planes ; and 3rd, that the usual assumed direction 
of the light is such, that its projections make angles of 45° with the 
ground line. 

165. The direction of the light itself, corresponding to the direc- 



SHADOWS. 61 

tion of its projections just mentioned, may be understood from an 
inspection of PI. VIII., Fig. 80. 

Let a cube be placed so that one of its faces, L'L L &, shall coincide 
with the vertical plane, and another face, I/aLi, with the horizontal 
plane. The diagonals, L'L t and I/I^, of these faces will be the projec- 
tions of a ray of light, and the diagonal, LL,, of the cube will be the 
ray itself; for the point of which 1/ and 1/ are the projections must 
be in each of the projecting perpendiculars L'L and I/L ; hence at 
L, their intersection. 

166. Referring now more particularly to the practical rules above 
mentioned (Art. 164), it appears — 1st, that as two points determine 
a straight line, we can determine the shadow of a straight line by 
passing rays through any two of its points, and finding where those 
rays pierce the surface receiving the shadow ; 2nd, that as one 
point will determine a line whose direction is already known, we 
Lave only to operate as above with a single ray, when the line 
casting the shadow is parallel, or perpendicular, to the plane receiv- 
ing the shadow. 

167. The essential facts relative to shadoios themselves, having 
been stated and illustrated in the foregoing articles, the object of 
this article is, to present, in a single view, those few and simple 
principles by which the projections of many ordinary shadows are 
constructed. 

Principle 1°. If a point A, Fig. 1, is in either plane of projec- 
tion, its projection on the other plane is on the ground line GL ; 
and its projection upon the plane containing it is only the point 
itself. Thus a' is the vertical projection of A-a. 

Principle 2°. If a point, A, Fig. 2, is on any line, as BC, its 
projections, a and a', will be on the projections, be and b'c', of that 
line. 

Principle 3°. The two projections of the same point, are always 
in the same perpendicular to the ground line (15) and (PI. I., Fig. 
3). See also aa' , Fig. 4. 

Now, recollecting that the shadow of a point upon a surface 
(162, 1st), is the intersection of a ray of light, through that point, 
with the given surface; we can, by the principles just given, find 
where the ray of light through any point pierces the planes of pror 
jection, or any other planes parallel to them ; the point, and the 
ray of light, being given by their projections. 

Thus, to find where the line or ray, AB (ab — a'b') Figs. 3 and 4, 
pierces the horizontal plane. (The inctorial construction, Fig. 3, 
and the actual one, Fig. 4, are similarly lettered.) 



62 



SHADOWS. 




Fig. 1. 



Fig. 2. 




Fig. 8. 



Fig. 4 



1st. The required point being in the horizontal plane, its verti- 
cal projection is (by Prin. 1°) in the ground line, GL. 

2nd. The point being on the given line, its vertical projection is, 
by (Prin. 2°), in the vertical projection of that line. 

Hence b\ the intersection of «'&'withGL,is the vertical projection 
of the intersection of ab — a'b' with the horizontal plane. The hori- 
zontal projection of b' , viz. b, which is also the required point itself 
(Prin. 1 Q ) is by (Prin. 2°) in ab, and, by (Prin. 3°), in the perpen- 
dicular, b'b\ to the ground line at b' . That is, b is the intersection 
of b'b with ab. 

The student, in advancing through the following problems, will 
now be prepared to note the point b, PL VIII., Fig. 83 ; the point 
b, Fig. 84 ; the point d, Fig. 85, and the points p and g, Fig. 88 ; 
etc. ; as illustrations of the foregoing construction. 



SHADOWS. 



63 



See also at this point, Figs. 5 and 6, which are exactly similar 
constructions of the point b\ where a line, or ray, AB (ab — a'b') 




Fig. 5. 




Fig. 6. 

pierces the vertical plane ; b' being the intersection of bb\ a per- 
pendicular to the ground line from b, where the horizontal projec- 
tion ab intersects it ; and a'b', the vertical projection of the given 
line. And the point b, Fig. 81 ; the points b and d, Fig. 82 ; the 
point O', Fig. 86 ; the points o\ h', g', etc., Fig. 87, etc., will be 
found to be examples of this construction. 

§ II. — Problems. 
168. Prob. 1. To find the shadow of a vertical beam, upon a 
vertical wall. PI. VIII., Fig. 81. Let AA' be the beam, let the 



64 



SHADOWS. 



vertical plane of projection be taken as the vertical wall, and let the 
light be indicated by the lines, as ab—a"b'. The edges which cast 
the visible shadow are a— a' a" ; ac—a" ; and ce—a"e'. The sha- 
dow of a — a' a" is a vertical line from the point b' , which point is 
where the ray from a— a" pierces the vertical plane. ab—a"b f 
pierces the vertical plane in a point whose horizontal projection is 
b. b' must be in a perpendicular to the ground line from b (Art. 15), 
and also in the vertical projection, a"b\ of the ray, hence at b'. b'b 
is therefore the shadow of a— a" a'. The shadow of ac— a" is the 
line b'd', limited by d', the shadow of the point c—a" (162). The 
shadow of ce—a"e' begins at d', and is parallel to ce— a"e', but is 
partly hidden. 

169. Execution. — The boundary of a shadow being determined, 
its surface is, in practice, indicated by shading, either with a tint of 
indian ink, or by parallel shade lines. The latter method, affording 
useful pen practice, may be profitably adopted. 

170. Pkob. 2. To find the shadow of an oblique timber, which is 
parallel to the vertical plane, upon a similar timber resting against 
the back of it. PI. VIIL, Fig. 82. Let AA' be the timber which 
casts a shadow on BB', which slants in an opposite direction. The 
edge ac—a'c' of AA', casts a shadow, parallel to itself, on the front 
face of BB'; hence but one fx>int of this shadow need be con- 
structed. Two, however, are found, one being a check upon the 
other. Any point, aa', taken at pleasure in the edge ac—a'c' casts 
a point of shadow on the front plane of BB', whose horizontal 
projection is b, and whose vertical projection (see Prob. 1) must be 
in the ray a'b', and in a perpendicular to the ground line at b ; 
hence it is at b'. The shadow of ac — a'c' being parallel to that 
line, b'd' is the line of shadow. d', the shadow of c — c', was found 
in a similar manner to that just described. 

It makes no difference that b' is not on the actual timber, BB'; 
for the face of that timber is but a limited physical plane, forming 
a portion of the indefinite immaterial plane, in which bb' is found ; 
hence the point b' is as good for finding the direction of the indefi- 
nite line of shadow, b'd', as is d', on the timber BB', for finding the 
real portion only of the line of shadow, viz. the part which lies 
across BB'. 

Observe that the back upper edge of AA', which is a heavy line, 
as seen in the plan, is so represented only till the intersection, near 
b — cd of the two timbers. 



PL.YI 



\ — ' 




i ! 




B 




B 


,L ! 


A'" 








V s 








JS. 


U 








SHADOWS. 



65 



171. Prob. 3. To find the shadow of a fragment of a horizon- 
tal timber, upon the horizontal top of an abutment on which the 
timber rests. PI. VIII., Fig. 83. Let AA' be the timber, and BB' 
the abutment. The vertical edge, a — a'V, casts a shadow, ab, in 
the direction of the horizontal projection of a ray (162, 3rd), and 
limited by the shadow of the point a— a'. The shadow of a— a' is 
at bb', where the ray ab—a'V pierces the top of the abutment ; b' 
being evidently the vertical projection of this point, and b being 
both in a perpendicular, b% to the ground line and in ab, the hori- 
zontal projection of the ray. The shadow of ad— at is bb", parallel 
t0 ad— a', and limited by the ray a'U—db". The shadow of 
de—a'e' is 5"c, parallel to de—a'e', and limited by the edge of the 
abutment. 

172. Prob. 4. To find the shadow of an oblique timber, upon a 
horizontal timber into which it is framed. PL VIII , Fig. 84. 
The upper back edge, ca— c'a\ and the lower front edge through 
ee\ of the oblique piece, are those which cast shadows. By con- 
sidering the point, c, in the shadow of be, Fig. 78, it appears that 
the shadow of ac—a'c', Fig. 84, begins at cc', where that edge 
pierces the upper surface of the timber, BB', which receives the 
shadow. Any other point, as aa', casts a shadow, W, on the plane 
of the upper surface of BB', whose vertical projection is evidently 
b\ the intersection of the vertical projection, a'b\ of the ray ab—a f U 
and the vertical projection, eV, of the upper surface of BB', and whose 
horizontal projection, b, must be in a projecting line, b% and in the 
horizontal projection, ah, of the ray. Likewise the line through ee\ 
and parallel to eb, is the shadow of the lower front edge of the 
oblique timber upon the top of BB'. This shadow is real, only so 
far as it is actually on the top surface of BB', and is visible and 
therefore shaded, only where not hidden by the oblique piece. 
Where thus hidden, its boundary is clotted, as shown at ea. The 
point, bb\ is in the plane of the top surface of BB', produced. 

HemarJc.— Thus it appears that when a line is oblique to a plane 
containing its shadow, the direction of the shadow is unknown till 
found. 

173. Prob. 5. To find the shadow of the side wall of a flight of 
steps upon the faces of the steps. PI. VIIL, Fig. 85. The steps 
can be easily constructed in good proportion, without measure- 
ments, by making the height of each step two-thirds of its width, 
taking four steps, and making the side walls rectangular parallelo- 
pipedons. 



66 SHADOWS. 

The edges, aa" — a! and a — ra', of the left hand side wall are 
those which cast shadows on the steps. 

The former line casts horizontal shadows parallel to itself, on 
the tops of the steps (162, 2nd), and shadows on the fronts of the 
steps in the direction of the vertical projection, a'd', of a ray of 
light (162, 3d) — from the upper step down to the shadow of the 
point aa'. Likewise, the edge a — ra! casts vertical shadows on 
the fronts of the steps (160), and shadows on their tops, parallel to 
ad, the horizontal projection of a ray (162, Srd) — from the lower 
step, up to the* shadow of aa' ; which is therefore the principal 
point to be found. In finding the shadow of aa', we have to 
remember that the vertical projection of any horizontal plane, as 1, 
is a straight line, b'l, and, if a point, as b, is in that plane, its 
vertical projection, b', will be in this line. Likewise that the 
horizontal projection of a vertical plane, as c'd'2$, is a straight 
line, cd", and if a point, c', is in that plane, its horizontal projec- 
tion, c, will be in this line. Accordingly, we readily find by 
inspection, or simple trial, that the ray of light, ad — a'd', in the 
present case, pierces the top surface of the third step at dd', where 
d' is found by projecting d' into ad. 

For the rest, the shadow cc", for example, is found by projecting 
down c'y and h'g', for example, is found by projecting up g. 

Let the student, ^to make a special exercise, vary the proportion 
of the steps or the direction of the light, so as to bring the shadow 
of aa' upon the vertical face of a step. 

174. Prob. 6. To find the shadow of a short cylinder, or washer, 
upon the vertical face of a board. PI. VIII., Fig. 86. Since the 
circular face of the washer is parallel to the vertical face of the 
board, BB', its shadow will be an equal circle (163), of which we 
have only to find the centre, 00'. This point will be the shadow 
of the point CC of the washer, and is where the ray CO — CO' 
pierces the board BB'. The elements, rv — r' and tu — t', of the 
cylindrical surface have the tangents r'r" and ft" for their shadows. 
These tangents, with the semicircle t"r", make the complete outline 
of the required shadow. 

175. Prob. 7. To find the shadow of a nut, upon a vertical 
surface, the nut having any position. PI. VIII., Fig. 86. Let 
a'c'e' — ace be the projections of the nut, and BB'the projections 
of the surface receiving the shadow. The edges, «V — ac and 
ce — c'e', of the nut cast shadows parallel to themselves, since they 
are parallel to the surface which receives the shadow. aV — an 



SHADOWS. 67 

are the two projections of the ray which determines the joint of 
shadow, nn' ; c'o' — co are the projections of the ray used in find- 
ing oo f , and e'r' — er is the ray which gives the point of shadow, 
rr f . The edges at aa! and ee', which are perpendicular to BB', cast 
shadows a'^'and e'r', in the direction of the projection of a ray of 
light on BB'. (See cb', the shadow of cb, PL VIII., Fig. 78.) 

176. Prob. 8. To find the shadow of a vertical cylinder, on a 
vertical plane. PL VIIL, Fig. 87. The lines of the cylinder, CC, 
w T hich cast visible shadows, are the element a — a" a', to which the 
rays of light are tangent, and a part of the upper base. The 
shadow of a — a n a\ is gg', found by the method given in Prob. 1. 
At g', the curved shadow of the upper base begins. This is found 
by means of the shadows of several points, bd', cc', dd f , &c. Each 
of these points of shadow is found as g was, and then they are 
connected by hand, or by the aid of the curved ruler. 

It is well to construct one invisible point, as w', of the shadow, to 
assist in locating more accurately the visible portion of the curved 
shadow line. 

177. Peob. 9. To find the shadow of a horizontal beam, upon 
the slanting face of an oblique abutment. PL VIII., Fig. 88. The 
simple facts illustrated by PL VIIL, Figs. 78-79, have no reference 
to the case of surfaces of* shadow, other than vertical or horizontal. 
But they illustrate the fact that the point where a line (see be, 
Fig. 78) pierces a surface, is a point of the shadow of the line 
upon such surface. To solve this problem, therefore, in a purely 
elementary manner, we will proceed indirectly, i. e., by finding the 
shadow on the horizontal top of the abutment and on its horizontal 
base. The point where these shadows meet the front edges of 
this top and this base, will be points of the shadow on the slanting 
face, qne. (See a', Fig. 78.) 

By Prob. 3 is found gc, the shadow of the upper back edge, 
ax — a'x', of the timber, AA', upon the top of the abutment, c, the 
point where it meets the front edge, ec, is a point of the shadow of 
AA' on the inclined face. By a similar construction with any ray, 
as bp — b'p', is found qp, the shadow of c'b' — db upon the base of 
the abutment; and q, where it intersects nq, is a point of the shadow 
of db — c'b' on the face, qne. The point, dd', where the edge, 
db — c'b', meets dc, is another point of the shadow of that edge ; 
hence dq — d'q f is the shadow of the front lower edge, db — c'b', on 
the iuclined face of the abutment. The line through cc', parallel to 
dq — d'q', is the shadow of the upper back edge, ax — a'x', and com- 
pletes the solution. 



68 SHADOWS. 

178. Prob. 10. To find the shadow of a pair of horizontal Um- 
bers, which are inclined to the vertical plane, upon that plane. PL 
VIII., Fig. 89. Let the given bodies be situated as shown in the 
diagram. In the elevation we see the thickness of one timber only, 
because the two timbers are supposed to be of equal thickness and 
halved together. As neither of the pieces is either parallel or 
perpendicular to the vertical plane, we do not know, in advance, 
the direction of their shadows. It will therefore be necessary to 
find the shadows of two points of one edge, and one, of the diago- 
nally opposite edge, of each timber. The edges which cast shadows 
are ac — aV and ht — e'h', of one timber, and ed — e'h and mv — a'm" 
of the other. All this being understood, it will be enough to point 
out the shadows of the required points, without describing their 
construction. (See bb', Fig. 81.) bb' is the shadow of aa', and dd f 
is the shadow of cc' ; hence the shadow line b' — d is determined. 
So uu' is the shadow of hh' ; hence the shadow line u'w' may be 
drawn parallel to b'd' . Similarly, for the other timber, ff is the 
shadow of ee', and oo' of mm'. One other point is necessary, which 
the student can construct. The process might be shortened some- 
what by finding the shadows of the points of intersection, p andr, 
which would have been the points//' and r" of the intersection of 
the shadows, and thus, points common to both shadows. 

1 79. Peob. 11. To find the shadow of a pair of horizontal Um- 
bers, which intersect as in the last problem, upon the inclined face 
of an oblique abutment. PL VIII., Fig. 90. The opening remarks 
of Prob. 9 apply equally well to this present one ; and the methods 
of solution being the same in both, only the peculiar features of 
the present problem will be noticed. Since the timber, BB', lies 
in the same direction with the light, the intersection of its two ver- 
tical faces with the top of the abutment are its shadows on that 
top, and oo' and vv', where those shadows meet en — c'n', the front 
edge of the abutment, are points of the shadow on the longer 
oblique face of the abutment. The points r p and t', on the lower 
edge of the same slanting face, are similarly found, and o'v'r'f 
is the required shadow of BB'. 

The shadow line, en, of the piece AA' on the top of the abutment 
is found as was gc in Fig. 88. n — n' is a point of the shadow 
of AA' on the inclined face of the abutment. It is necessary to find 
two points of the shadow of its lower front edge, be — b'c'. cc' is 
one of these points, being the point where the edge be — b'c' meets 
the edge en — c'n' of the abutment. In attempting to find a point 



SHADOWS. 60 

of shadow on the lower edge, £V, of the abutment by finding an 
indefinite shadow line, like pq, Fig. 88, on its lower base, we meet 
with the inconvenience of finding this shadow to be very far out 
of the central portion of the figure, so that a point, as q, in Fig. 88, 
will n//t be below the ground line. 

To meet this case, take an intermediate horizontal plane, as 
s'd', and find its intersection, s'd' — sq, with the abutment, by pro- 
jecting down the intersection of s'd' with any corner edge, upon 
the plan of the same edge, and by drawing through the point so 
found parallels, as sq, to the lower edges of the abutment. Now, 
bb' is a point of the lower front edge of AA'. The ray bd — b'd\ 
drawn through it, pierces the plane s'd' at d', whose horizontal 
projection is in rd at d ; and ds is the indefinite shadow of the lower 
front edge of the timber on the same plane ; then ss' is a point in 
the required shadow on the inclined face of the abutment; and by 
drawing c's' and n'p, parallel to it, we have the required shadow in 
vertical projection. The horizontal projection of the shadow is 
omitted, to avoid confusing the diagram. It is contained between 
a line joining c and s, and a line through n, parallel to cs. 

180. Prob. 12. To find the shadoio of the floor of a bridge 
upon a vertical cylindrical abutment. PI. VIII., Fig. 91. The 
line ag — a'g' is the edge of the floor which casts the shadow. 
bdg — b'e'g'n is the concave vertical abutment receiving the shadow. 
gg', where the edge ag — g'a' of the floor meets this curved wall, is 
one point of the shadow, fis the horizontal projection of the point 
where the ray, ef — e'f, meets the abutment ; its vertical projection 
is in e'f, the vertical projection of the ray, and in a perpendicular 
to the ground line, through f hence at/*'. Similarly we find the 
points of shadow, dd' and bb\ and joining them with/' and g\ have 
the boundary of the required shadow. Observe, that to find the 
shadow on any particular vertical line, as b — b"b\ we draw the ray 
in the direction b — a ; then project a at a', &c. 

Remark. — The student may profitably exercise himself in chang- 
ing the positions of the given parts, while retaining the methods 
of solution now given. 

For example, let the parts of the last problem be placed side 
by side, as two elevations, giving the shadow of a vertical wall 
on a horizontal concave cylindrical surface ; or, let the timbers, Fig. 
89, be in vertical planes, and let their shadows then be found on a 
horizontal surface. 



CHAPTER n. 

SHADING. 

181. Notes. — 1. It is not proposed here to go largely into the 
theory of shading in general, but to make a special study of each 
one of the four practically useful exercises here given, and to notice 
only such principles as are involved in their elaboration. 

2. The general problem for the author has been, on the one hand, 
to give nothing " by rote," or arbitrarily, but rather on rational 
grounds ; and on the other hand, to discover methods so elementary, 
and solutions so simple in form, as to render the problems of this 
Division of the course perfectly intelligible to the student at this 
early stage of his general course of graphical study. 

182. Example 1°. To shade the elevation of a vertical right 
hexagonal prism, and its shadow on the horizontal plane. PL 
IX., Fig. 92. Let the prism be placed as represented, at some dis- 
tance from the vertical plane, and with none of its vertical faces 
parallel to the vertical plane. The face, A, of the prism is in the 
light ; in fact, the light strikes it nearly perpendicularly, as may be 
seen by reference to the plan ; hence it should receive a very light 
tint of indian ink. The left hand portion of the face, A, is made 
slightly darker than the right hand part, it being more distant; 
for the reflected rays, which reach us from the left hand portion, 
have to traverse a greater extent of air than those from the neigh- 
borhood of the line tt\ and hence are more absorbed or retarded; 
or, without attempting to say what happens to these rays, the fact 
is, that they make a weaker impression on the eye, causing the left 
hand portion of A, from which they come, to appear darker than 
at W. 

183. Remarks. — a. It should be remembered that the whole of 
face A is very light, and the difference in tint between its opposite 
sides very slight. 

b. As corollaries from the preceding, it appears : first, that a sur- 
face parallel to the vertical plane would receive a uniform tint 
throughout ; and, second, that of a series of such surfaces, all of 



SHADING. 11 

which are in the light, the one nearest the eye would be lightest, 
and the one furthest from the eye, darkest. 

c. It is only for great differences in distance that the above effects 
are manifest in nature ; but the whole system of drawing by projec- 
tions being an artificial one, both in respect to the shapes which it 
gives in the drawings, and in the absence of surrounding objects 
which it allows, we are obliged to exaggerate natural appear- 
ances in some respects, in order to convey a clearer idea of the forms 
of bodies. 

d. The mere manual process of shading small surfaces is here 
briefly described. With a sharp-pointed camel's-hair brush, wet 
with a very light tint of indian ink, make a narrow strip against the 
left hand line of A, and soften off its edge with another brush 
slightly wet with clear water. When all this is dry, commence at 
the same line, and make a similar but wider strip, and so proceed 
till the whole of face A is completed, when any little irregularities 
in the gradation of shade can be evened up with a delicate sable 
brush, damp only with very light ink. 

184. Passing now to face B, it is, as a whole, a little darker than 
A, because, as may be seen by reference to the plan, while a beam 
of rays of the thickness mp strikes face A, a beam having only a thick- 
ness pv, strikes i'ace B ; i. e., we assume, first, that the actual bright- 
ness of a flat surface is proportioned to the number of rays of light 
which it receives; and, second, that its apparent brightness is, other 
things being the same, proportioned to its actual brightness. Also, 
the part at a — a' being a little more remote than the line t — 1\ the 
part at a — a' is made a very little darker. 

185. The face C is decidedly darker than B, as it receives no light, 
except the small amount which it receives by reflection from sur- 
rounding objects. This side, C, is darkest at the edge a — a' which 
is nearest to the eye. This statement is confirmed by experience ; 
for while the shady side of a house near to us appears in strong con- 
trast with the illuminated side, the shady side of a remote building 
appears scarcely darker than the illuminated side. This fact is 
explained as follows : The air, together with the various floating 
particles in it, is, partly, a reflecting medium, intensely illuminated 
by the sun. Some of its reflections enter the eye in the direction 
of the shady surface, and the eye attributes to that surface all the 
light which comes in that direction. But, the more distant the 
body, the greater will be the amount of these atmospheric reflec- 
tions coming in the same direction with the very few rays remitted 
by the shady surface itself. Hence the more distant that surface, the 



12 SHADING. 

lighter it will appear. It may be objected that this would make out 
the remoter parts of illuminated surfaces as the lighter parts. But 
not so ; for the air is a nearly perfect transparent medium, and hence 
reflects but little, compared with what it transmits to the opaque 
body ; but being not quite transparent, it absorbs the reflected rays 
from the distant body, in proportion to the distance of that body, 
making therefore the remoter portions darker; while the very wea7c 
reflections from the shady side are reinforced or replaced by more 
of the comparatively stronger atmospheric reflections, in case of the 
remote, than in case of the near part of that shady side. Thus is 
made out a consistent theory. 

In relation, now, to the shadow, it will be lightest where furthest 
from the prism, since the atmospheric reflections evidently have to 
traverse a less depth of darkened air in the vicinity of de — d' than 
near the lower base of the prism at abc. 

186. Ex. 2°. To shade the elevation of a vertical cyli?ider. PI. IX., 
Fig. 93. Let the cylinder stand on the horizontal plane. The figures 
on the elevation suggest the comparative depth of color between 
the lines adjacent to the figures. The reasons for so distributing 
the tints will now be given. 

The darkest part of the figure may properly be assumed to be 
that to which the rays of light are tangent ; viz., the vertical line at 
tt\ from which the tint becomes lighter in both directions. 

The lightest line will be at the place which reflects the most light 
to the eye. Now it is a principle of optics that the incident and 
the reflected ray make equal angles with a perpendicular to a sur- 
face. But nQ is the incident and Ce the reflected ray, to the eye ; 
hence at d is the point, on the surface, where they make equal 
angles with the perpendicular (normal) dC, and the vertical line at 
d — d' is the lightest line. 

187. HemarJc. — The question may here arise, "If all the light 
that is reflected towards the eye is reflected from d — as it appears 
to be — how can any other point of the body be seen ?" To answer 
this question requires a notice— -first, of the difference between 
polished and dull surfaces ; and, second, between the case of light 
coming wholly m. one direction, or principally in one direction. If 
the cylinder CO', considered as perfectly polished, were deprived of 
all reflected light from the air and surrounding objects, the line at 
d — d' would reflect to the eye all the light that the body would 
remit towards the eye, and would appear as a line of brilliant light, 
while other parts remitting no light whatever would be totally 



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SHADING. 73 

invisible. Let us now suppose a reflecting medium, though an 
imperfect one, as the atmosphere, to be thrown around the body. 
By reflection, every part of the body would receive some light from 
all directions, and so would remit some light to the eye, making the 
body visible, though faintly so. But no body has a polish that is 
absolutely perfect ; rather, the great majority of those met with in 
engineering art have entirely dull surfaces. Now a dull surface, 
greatly magnified, may be supposed to have a structure like that 
shown in PI. IX., Fig. 97, in which many of the asperities may be 
supposed to have one little facet each, so situated as to remit to the 
eye a ray received by the body directly from the principal source 
of light. 

188. Having thus proved that a cylinder placed before our eyes 
can be seen, we may proceed with an explanation of the distribu- 
tion of tints, b is midway between d and t. At b, the ink may be 
diluted, and at e — e', much more diluted, as the gradation from a 
faint tint at e to absolute whiteness at d should be without any 
abrupt transition anywhere. 

The beam of incident rays which falls on the segment dn, is 
broader than that which strikes the equal segment de / hence the 
segment nd is, on the elevation, marked 5, as being the lightest band 
which is tinted at all. The segment nr, being a little more obliquely 
illuminated, is less bright, and in elevation is marked 3, and may be 
made darkest at the left hand limit. Finally, the segment rv receives 
about as much light as rn, but reflects it within the very narrow 
limits, s, hence appears brighter. This condensed beam, s, of 
reflected rays would make rv the lightest band on the cylinder, but 
for two reasons ; first, on account of the exaggerated effect allowed 
to increase of distance from the eye ; and, second, because some of 
the asperities, Fig. 97, would obscure some of the reflected rays 
from asperities still more remote ; hence rv is, in elevation, marked 
4, and should be darkest at its right hand limit. 

189. The process of shading is the same as in the last exercise. 
Each stripe of the preliminary process may extend past the preced- 
ing one, a distance equal to that indicated by the short dashes at 
the top of Fig. 94. When the whole is finished, there should be a 
uniform gradation of shade from the darkest to the lightest line, 
free from all sudden transitions and minor irregularities. 

190. Ex. 3°. To shade a right cone standing upon the horizon- 
tal plane, together with its shadow. PI., IX., Fig. 95. The sha- 
dow of the cone on the horizontal plane, will evidently be bounded 



74 SHADING. 

by the shadows of those lines of the convex surface, at which the 
light is tangent. The vertex is common to both these lines, and 
casts a shadow, v'". The shadows being cast by straight lines of 
the conic surface, are straight, and their extremities must be in the 
base, being cast by lines of the cone, which meet the horizontal 
plane in the cone's base; hence the tangents v'"t and v"'t" are thi 
boundaries of the cone's shadow on the horizontal plane, and the 
lines joining t' and t" with the vertex are the lines to which tho 
rays of light are tangent ; i. e., they are the darkest lines of the 
shading ; hence tv, the visible one in elevation, is to be vertically 
projected at t'v'. 

The lightest line passes from vv' to the middle point, y, between 
n and p in the base. At q and at p a change in the darkness of the 
tint is made, as indicated by the figures seen in the elevation. In 
the case of the cone, it will be observed that the various bands of 
color are triangular rather than rectangular, as in the cylinder ; so 
that great care must be taken to avoid filling up the whole of the 
upper part of the elevation with a dark shade. 

191. Ex. 4°. To shade the elevation of a sphere. PI. IX., Fig. 
96. It is evident that the parallel tangents to a sphere, at all the 
points of a great circle, will be a system of parallel lines which will 
be perpendicular to that circle. Hence, all the space around the 
sphere being filled with parallel rays, some ray will be found tan- 
gent to the sphere at each point of a great circle, perpendicular to 
those rays. Such a great circle will be the curve of darkest shade 
on the sphere. Next, let us consider, that every circle, whether 
great or small, which is cut from the sphere in the same direction 
as that in which the light proceeds, will have two rays tangent to 
it on opposite sides, whose points of tangency must be points of the 
circle of shade ; being points in which rays of light are tangent to 
the surface of the sphere. 

Next, let us recollect that always, when a line is parallel to a plane, 
its projection on that plane will be seen in its true direction. Now 
BD' being the direction of the light, as seen in elevation, let BD' be 
the trace, on the vertical plane, of projection — taken through the 
centre of the sphere — of a new plane perpendicular to the vertical 
plane, and therefore parallel to the rays of light. The projection 
of a ray of light on this plane, BD', will be parallel to the ray itself, 
and therefore the angle made by this projection with the trace BD' 
will be equal to the angle made by the ray with the vertical plane. 
But, referring to PI, YIII., Fig. 80, we see that in the triangle 



SHADING. 75 

LL'L 1? containing the angle LLiL' made by the ray LL t with the ver- 
tical plane, the side LT^ is the hypothenuse of the triangle ~L'b L„ 
each of whose other sides is equal to LI/. Hence in PI. IX., 
Fig. 96, take any distance, Be, make AB perpendicular to BD', and 
on it lay off BD = Be, then make BD' = Dc, join D and D', and DD' 
will be the projection of a ray upon the plane BD', and BD'D will 
be the true size of the angle made by the light with the vertical 
plane; it being understood that the plane BD', though in space 
perpendicular to the vertical plane, is, in the figure, represented as 
revolved over towards the right till it coincides with the vertical 
plane of projection, and with the paper. 

192. We are now ready to find points in the curve of shade, oo' 
is the vertical projection of a small circle parallel to the plane BD', 
and also of its tangent rays. The circle og'o', on oo' as a diameter, 
represents the same circle revolved about oo' as an axis and into 
the vertical plane of projection. Drawing a tangent to og'o\ 
parallel to DD', we find g\ a visible point of the curve of shade, 
which, when the circle revolves back to the position oo ', appears at 
g, since, as the axis oo' is in the vertical plane, an arc, g'g, described 
about that axis, must be vertically projected as a straight line. (See 
Art. 39.) 

In a precisely similar manner are found A, 7c, m, and f. At A 
and B, rays are also evidently tangent to the sphere. Through 
A, f, &c, to B the visible portion of the curve of shade may now 
be sketched. 

193. The most highly illuminated point is 90° distant from the 
great circle of shade ; hence, on ABQ, the revolved position of a 
great circle which is perpendicular to the circle of shade, lay off 
^'Q = ^B=zthe chord of 90°, and revolve this perpendicular circle 
back to the position qq\ when Q will be found at r'. But the 
brilliant point, as it appears to the eye, is not the one which receives 
most light, but the one that reflects most, and this point is midway, 
in space, between r' and r, i. e. at P, found by bisecting QB, and 
drawing HP ; for at R the incident ray whose revolved position is 
Qr, parallel to DD', and the reflected ray whose revolved position 
is rB, make equal angles with R'R, the perpendicular (normal) to 
the surface of the sphere. 

194. In regard now to the second general division of the problem 
— the distribution of tints ; a small oval sj)ace around P should be 
left blank. The first stripe of dark tint reaches from A to B, along 
the curve of shade, and the successive stripes of the same tint extend 
to B^A on one side of B&A, and to oew on the other. Then take 



76 SHADING. 

a lighter tint on the lower half of the next zone, and a still lighter 
one on its upper half (2) and (3). In shading the next zone, use an 
intermediate tint (3-4), and in the zone next to P a very light tint on 
the lower side (4), and the lightest of all on the upper side (5). 
After laying on these preliminary tints, even up all sudden transi- 
tions and minor irregularities as in other cases. 



DIVISION FOURTH. 

ISOMETEICAL DRAWING, AND CABINET PROJECTIONS. 



CHAPTER I. 

FIRST PRINCIPLES OF ISO-METRICAL DRAWING. 

195. From what has been seen in the course on Projection? 
(Division First), it appears that when two lines are equal, and 
both parallel to the planes of projection, as in PI. I., Figs. 1, 2, their 
projections are equal. This has also frequently been illustrated in 
the problems of Division Second. 

196. It has also been seen, that while a line viewed perpendicu- 
larly, i. e., a line parallel to the planes of projection, is seen in its 
true size, — one which is viewed obliquely, appears shorter than it 
really is. This has been repeatedly shown by comparison of the 
two projections of the edges of a nut, in the problems of Division 
Second. 

197. From the foregoing, it is not difficult to see that if any two 
lines be equal and parallel to each other in space, their projections 
will also be equal and parallel to each other. This is easily p*roved, 
however ; for as the lines in space are equal and parallel, the dis- 
tance between the projecting lines of one of them is equal to the 
distance between the projecting lines of the other. Hence the pro- 
jections of equal and parallel lines are equal and parallel — to each 
other — but not to the lines in space, wken the latter are oblique to 
the planes of projection. 

198. But it is not the parallelism, but the equal inclination that 
makes the projections equal; hence equal lines, equally inclined 
to the planes of projection, will have equal projections. Conversely, 
if equal projections are the projections of equal lines, those lines 
must be equally inclined to the planes of projection. 



7 8 FIRST PRINCIPLES OF IS0METR1CAL DRAWING. 

199. To apply the foregoing principles to the development of the 
elements of Isometrical Drawing, let us take a cube, and begin by 
supposing that we see it in right vertical projection, as in PI. IX., 
Fig. 98. In this case, but one side of the cube will be visible, and 
that will appear as a square. 

Next, suppose the cube to be turned horizontally, until two of its 
vertical faces should become visible. Evidently it can be turned 
just so far as to show those two faces equally, as in PL IX., Fig. 99, 
where a — a' and b — b' are equal. 

If now, without turning the cube any more to the right or left, 
its upper back corner, dd\ be turned up towards the eye, it is plain 
that we should see the upper face in addition to the two faces seen 
in the last elevation, PL IX., Fig. 99. It is also plain that the cube 
may thus be turned up just far enough to make these three visible 
faces appear equal, as in PL IX., Fig. 100. Then the edge, c — c\ 
Fig. 99, will appear equal to Ce" and Of", Fig. 100. 

200. Let us now observe carefully the facts which result when 
the corner dd' is turned up just far enough to make the three visible 
faces of the cube appear equal. 

1st. When the faces of the cube appear equal, the sides of those 
faces, i. e., the edges of the cube must appear equal ; hence a — a\ 
b — b\ c — c\ Fig. 99, for example, will be equal in the new pro- 
jection. 

2nd. Being equal, both in projection and in space, these edges 
must be equally inclined to the plane of projection (198). 

3rd. These edges being equal, equally inclined to the plane of 
projection, and also equally inclined to each other — being at right 
angles to each other in space — their included angles will be equal in 
projection. (See PL III., Fig. 35, the plan.) Hence, as a first step 
in constructing the new projection, lay off from any assumed point, 
C, Pl.'IX., Fig. 100, three lines making angles of 120° with each 
other ; observing that if one, Cn, be vertical, the other two, Of" 
and Ce", will make angles of 30° with a horizontal line through C, 
and can therefore be easily drawn with the 30° triangle. 

4th. The diagonal ef—e'f of the upper base, PL IX., Fig. 99, 
remaining parallel to the vertical plane, will still be seen in its true 
size; hence from C, PL IX., Fig. 100, lay off half of the diagonal 
ef — e'f from Fig. 99, each way on the Kne EF, and at E and F erect 
perpendiculars, whose intersection with Ce" and Qf" will give the 
required diagonal in its true position, and consequently Ce" and C/*", 
two sides of the required projection, and projections of ce — d'e' and 
cf—d'f, Fig. 99. Make now Cn=Ce". 



FIRST PRINCIPLES OF ISO-METRICAL DRAWING. 1 9 

5th. All the other edges of the cube are equal and parallel to 0e'\ 
Of" , and On, They are therefore readily drawn, and the required 
projection completed. This projection is called the Isometrical 
Projection of the cube. 

6th. We see now, by inspection, that the perimeter, T5f"onpe\ of 
this projection is a regular hexagon, of whose circumscribing circle, 
C is the centre, and Oe", Of", and On, are equidistant radii. It 
also appears that Fig. 100 could have been constructed by taking 
e"f"-=ef from Fig. 99, as the side of an inscribed equilateral triangle, 
by then inscribing a hexagon in the circumscribing circle of that 
triangle, and by then drawing Ce", Of", and On. Again, the three 
face diagonals, e"f", f"n, and ne" , being homologous lines of equal 
and similarly placed faces, are all parallel to the plane of projection, 
and therefore equal in projection. Finally, the body diagonal of 
the cube, through C, is projected at 0, Fig. 102, the projection of 
the extreme front and back corners of the cube. But when a 
straight line is projected in a point, it is perpendicular to the plane 
of projection. Hence in isometrical projection, the plane of pro- 
jection is perpendicular to the body diagonal through the foremost 
corner of a cube. 

201. Summing up in the form of definitions : 

1st. Isometrical projection is that in which the three plane right 
angles, which form a solid right angle, appear equal in projection. 

2nd. The three lines meeting at C are called isometric axes. 

3rd. The point C is called the isometric centre. 

4:th. Planes containing any two of the isometric axes, or parallel 
to these axes, are called isometric planes. 

5 th. Lines parallel to the isometric axes are called isometric 
lines. 

202. It is to be noticed, that while it is the three diagonals only 
of the faces of the cube, Fig. 100, that are seen in their real size, or 
on the same scale as that used for the ordinary plan and elevation, 
yet the edges are the lines of which the measurements are usually 
given, and which it is therefore desirable to represent on the same 
scale in all its projections. If, therefore, a figure be constructed, as 
in PI. IX., Fig. 101, similar to Fig. 100, but having its edges, Da, 
&c, equal to the edges of the plan in Fig. 99, such a figure will be 
called an Isometrical Drawing of the cube shown in Fig. 99, while 
it would be the strict isometrical projection of a larger and imagi- 
nary cube. 

This imaginary cube is to Fig. 101, as Fig. 99 is to its projection 
in Fio\ 100. 



80 FIRST PRINCIPLES OP ISOMETRICAL DRAWING. 

203. Again, it is evident that it is the direction of the projection 
of the edges of a solid right angle which determines whether the 
projection is isometrical or not ; hence, not merely cubes, but any- 
rectangular objects, can readily be put into isometrical projec- 
tion. 

204. Recollecting that all lines of the object in space which are 
at right angles to each other, are shown in the isometrical drawing 
on the scale of the plan and elevation, and recollecting that in most 
pieces of carpentry, in boxes, furniture, &c, the principal lines are 
at right angles to each other, we see at once the great convenience 
of isometrical drawing, which shows the true dimensions of objects 
in three directions on the same figure, and gives a clearer idea of 
the construction of objects than plans and elevations do. The dis- 
agreeably distorted appearance, however, of large or complicated 
structures, as buildings or machines, when shown in isometrical 
projection, causes this method to be confined, in its application, to 
small objects, details, &c. 

205. It is often convenient to know in advance the extreme 
height of the isometrical drawing of a given object, in order that 
it may be better arranged upon the paper. See PI. X., Fig. 103, 
for example. Da being 10 inches — by scale — the distance of a 
below r a horizontal line through D, is the sine of 30° to a radius of 
10. Hence, as sin. 30°= J, that distance is 5 inches. In like man- 
ner aC being 4 inches, C is 2 inches below a ; and Cc being 3 J 
inches, c is 10J inches below D. 

205$. As a final preliminary, it is to be noted, that, as no planes 
of projection are commonly employed in isometrical drawing, the 
direction of the light is defined with respect to a cube as a standard 
body of reference. Thus, Fig. 102, the ray entering at a is sup- 
posed to leave the cube at^>, and all other rays are parallel to this. 



PL.VJII. 



f'fl'd'c 




CHAPTER n. 

PROBLEMS INVOLVING ONLY ISOMETRIC LINES. 

206. Prob. 1. To make the isometrical drawing of a cube, from 
a given plan and elevation. 

1st Method. The plan and elevation, PL IX., Fig. 98, repre- 
sents a cube, whose edge is one inch in length, seen in " right pro- 
jection.'''' To make the isometrical drawing of the same cube, Fig. 
101, take any point, C, as a centre, and describe a circle with one 
inch radius, CD ; lay off the radius six times, commencing at either 
end of the vertical diameter; join the points thus formed, and draw 
aC, bC, and cC. Then Ca and Ob, and all the lines parallel with 
them, are drawn with the 30° triangle. 

2nd Method. Draw the three axes, and lay off an inch on each 
of them. The remaining lines, being equal and parallel to these, 
are drawn with the 30° triangle, placed in different positions, which 
will readily suggest themselves, on attempting the construction. 

207. To complete the representation, certain lines are made heavy. 
If a line, LL', PL IX., Fig. 102, be drawn through the corner, a, of 
the cube, in the direction of a ray of light, it will pass through the 
cube and leave it at the corner, p. Those surfaces of the cube, 
therefore, which ore represented in Fig. 102 by CqDa, aCno, and 
aDCo, are illuminated, and equally so, because the diagonal, as ap, 
makes equal angles with all the faces of the cube. But Cqpn, C/j^D, 
and Cpno, are all in the dark; hence, according to the usual rule, 
that those lines which divide a dark from a light surface are made 
heavy, J)q, qC, On, and no, are all made heavy. 

Again, we know from elementary geometry that two intersecting 
lines fix the position of the plane containing them. Accordingly, 
one plane can contain oab and the ray LL'. This plane will also 
contain pq, parallel to oab, and hence will cut aDqO in aq. 

Therefore, rays through all points of ab will meet aDqC in aq. 
Hence, by drawing the ray bb', we find ab' , the shadow of ab, 
which is the edge oa produced. 

By similar reasoning, we find ad', the shadow of ad, which is the 
edge Da produced. 

208. Prob. 2. To represent a prismatic blocJc, as cut from a 



82 PROBLEMS INVOLVING ONLY ISOMETRIC LINES. 

corner of the cube. PI. IX., Fig. 101. This problem involves the 
construction of points in the isometric axes. Let the cube now 
be supposed to be one whose edge is five inches long, and let it be 
drawn to a scale of one fifth, or of one inch to five inches. Suppose 
a piece, each of whose sides is a rectangle, to be cut from the cor- 
ner nearest the eye. Let Ca'=2 inches, Cb r = 3 inches, and Cc'= 
1 inch. Lay off" these distances from C, upon the axes, and through 
the points a\ b\ and c', draw isometric lines, as a'<#and b'd, which, 
by their intersections, will complete the portion cut from the cube. 
Heavy lines are placed as in the last problem ; but all the lines of 
division between the body of the cube and the piece cut from it are 
light, according to the common rule, that when two surfaces are 
continuous and form one surface, the dividing line between them is 
fine ; and this, because such a dividing line does not separate a light 
from a dark surface, but merely lies in a light surface. 

209. Prob. 3. To construct the isometrical drawing of a carpen- 
ters oil stone box. PI. X., Fig. 103. This problem involves the 
finding of points which are in the planes of the isometric axes. 

Let the box containing the stone be 10 inches long, 4 inches wide? 
and 3 inches high, and let it be drawn on a scale of ^. 

Assume C then, and make Ca=4 inches, C#=10 inches, and 
Cc—3^ inches, and by other lines «D, D£, &c., equal and parallel to 
these, complete the outline of the box. 

Represent the joint between the cover and the box as being 1 inch 
below the top, aCb. Do this by making Cp=l inch, and through 
p drawing the isometric lines which represent the joint. 

Suppose, now, a piece of ivory 5 inches long and 1 inch wide to 
be inlaid in the longer side of the box. Bisect the lower edge at d, 
make de=l inch, and ee' = l inch. Through e and e\ draw the top and 
bottom lines of the ivory, making them 2|- inches long on each side 
of de\ as at e't, e'V. Draw the vertical lines at t and t\ which will 
complete the ivory. 

210. To show another way of representing a similar inlaid piece, 
let us suppose one to be in the top of the cover, 5 inches long and 1 j- 
inches wide. Draw the diagonals ab and CD, and through their 
intersection, o, draw isometric lines; lay off oo' =2j inches, and 
oo"—\ of an inch, and lay off equal distances in the opposite direc- 
tions on these centre lines. 

Through o\ </, &c, draw isometric lines to complete the repre- 
sentation of the inlaid piece. 

211. Prob. 4. To represent the same box (Prob. 3), with the cover 
removed, PI. X., Fig. 104. 



PROBLEMS INVOLVING ONLY ISOMETRIC LINES. 83 

This problem involves the finding of the positions of points not in 
the isometric planes. 

Supposing the edges of the box to be indicated by the same let- 
ters as are seen in Fig. 1 03, and supposing the body of the box to 
be drawn, 10 inches long, 4 inches wide, and 2j inches high; then, 
to find the nearest upper corner of the oil stone, lay off on Cb 1| 
inches, and through the point/, thus found, draw a line,^', paral- 
lel to Ca. On ff\ lay off 1 inch at each end, and from the points 
hh'i thus found, erect perpendiculars, as hn, each f of an inch long. 
Make the further end, a?, of the oil stone 1^ inches from the further 
end of the box, and then complete the oil stone as shown. To find 
the panel in the side of the box, lay off 2 inches from each end of the 
box, on its lower edge ; at the points thus found, erect perpendi- 
culars, of half an inch in length, to the lower corners, as^>, of the 
panel; make the panels 1^-. inches wide, as at pp\ and \ an inch 
deep, as at pr, which last line is parallel to Ca ; and, with the iso- 
metric lines through r,p,p\ and t, completes the panel. 

212. The manner of shading this figure will now be explained. 

The top of the box and stone is lightest. Their ends are a trifle 
darker, since they receive less of the light which is diffused through 
the atmosphere. The shadow of the oil stone on the top of the box 
is much darker than the surfaces just mentioned. The shadow of the 
foremost vertical edge of the stone is found in precisely the same way 
as was the shadow of the wire upon the top of the cube, PI. IX., Fig. 
102. The sides of the oil stone and of the box, which are in the dark, 
are a little darker than the shadow, and all the surfaces of the panel 
are of equal darkness and a trifle darker than the other dark surfaces. 
In order to distinguish the separate faces of the panel, when they are 
of the same darkness, leave their edges very light. The little light 
which those edges receive is mostly perpendicular to them, regard- 
ing them as rounded and polished by use. These light lines are left 
by tinting each surface of the panel, separately, with a small brush, 
leaving the blank edges, which may, if necessary, be afterwards 
made perfectly straight by inking them with a light tint. The 
tipper and left hand edges of the panel, and all the lines correspond- 
ing to those which are heavy in the previous figures, may be ruled 
with a dark tint. In the absence of an engraved copy, the figures 
will indicate tolerably the relative darkness of the different surfaces; 
1 being the lightest, and the numbers not being consecutive, so that 
they may assist in denoting relative differences of tint. When this 
figure is thus shaded, its edges should not be inked with ruled linea 
in black ink, but should be inked with pale inl? 



CHAPTER HI. 

PROBLEMS INVOLVING NON-ISOMETKICAL LINES. 

213. When non-isometrical lines occur, each point in any of them 
which is in an isometric plane, must be located by two isometric 
lines representing lines which in common projection would be found 
at right angles to each other, as is illustrated in the following pro- 
blems — which, by the way, are numbered continuously through the 
chapters of this Division. 

214. Peob. 5. To construct the isometrical drawing of the 
scarfed splice, shown at PI. VII., Fig. 66. Let the scale be T V, or 
three fourths of an inch to a foot. In this case, PL X., Fig. 105, it 
will be necessary to reconstruct a portion of the elevation to the 
new scale (see PI. X., Fig. 106), where Ap=l% feet, jt?A ff =6 inches, 
and the proportions and arrangement of parts are like PL VII., 
Fig. 66. In Fig. 105, draw AD, 3 feet; make DB and AA' each 
one foot, and draw through A' andB isometric lines parallel to AD. 
Join A — B, and from A lay off distances to 1, 2, 3, 4, equal to the 
corresponding distances on Fig. 106. Also lay off, on BK, the same 
distances from B, and at the points thus found on the edges of the 
timber, draw vertical isometric lines equal in length to those which 
locate the corners of the key in Fig. 106. Notice that opposite 
sides of the keys are parallel, and that AV, and its parallel at B, 
are both parallel to those sides of the keys which are in space per- 
pendicular to AB. To represent the obtuse end of the upper tim. 
berof the splice, bisect AA', and make va=Aa, Fig, 106, and draw 
Aa and A' a. Locate m by va produced, as at ar, Fig. 106, and a 
short perpendicular=rm, Fig. 106, and draw mV and mV; W 
being parallel to AA', and am being parallel to AV. To represent 
tjie washer, nut, and bolt, draw a centre line, vv\ and at t, the mid- 
dle point of V/i, draw the isometric lines tu and ue, which will give 
e, the centre of the bolt hole or of the bottom of the washer. A point 
^coinciding in the drawing with the upper front corner of the nut — 
is the centre pf the top of the washer, which may be ma.de | of an 
inch thick. 



PROBLEMS INVOLVING NON-ISOMETEICAL LINES. 85 

Through the above point draw isometric lines, rr' and pp', and 
lay off on them, from the same point, the radius of the washer, say 
2^- inches, giving four points, as o, through which, if an isometric 
square be drawn, the top circle of the washer can be sketched in it, 
being tangent to the sides of this square at the points, as o, and 
elliptical (oval) in shape. The bottom circle of the washer is seen 
throughout a semi-circumference, i. e. till limited by vertical tan- 
gents to the upper curve. 

On the same centre lines, lay off from their intersection, the half 
side of the nut, 1 <J } inches, and from the three corners which will be 
visible when the nut is drawn, lay off on vertical lines its thickness, 
li inches, giving the upper corners, of which c is one. So much 
being done, the nut is easily finished, and the little fragment of bolt 
projecting through it can be sketched in. 

The other nut, being similarly constructed, is omitted. The nut 
is here constructed in the simplest position, i.e. with its sides in the 
direction of isometric lines. If it had been determined to construct 
it in any oblique position, it would have been necessary to have 
constructed a portion of the plan of the timber with a plan of the 
nut — then to have circumscribed the plan of the nut by a square, 
parallel to the sides of the timber — then to have located the cor- 
ners of the nut in the sides of the isometric drawing of the circum- 
scribed square. Let the student draw the other nut on a large scale 
and in some such irregular position. See Fig. 107, where the upper 
6gure is the isometrical drawing of a square, as the top of a nut ; 
this nut having its sides oblique to the edges of the timber, which 
are supposed to be parallel to ca. 

215. Peob. 6. To make an isometrical drawing of an oblique 
timber framed into a horizontal one. PI. X., Fig. 108. Let the 
original be a model in which the horizontal piece is one inch square, 
and let the scale be ±. 

Make ag and ah, each one inch, complete the isometric end of the 
horizontal piece, and draw ad, hn, and gJc. On ad, lay off ab—2 
inches, and draw bm, bc=2± inches, and cd=2 inches. Make 
de=% of an inch, draw ce, and lines through b and m parallel to it. 
Let /be } of an inch below ad and 2| inches from ag, measured 
from a on ad, and draw bf and fc, which, with the representation 
of the broken ends, will complete the figure. 

216. Peob. 7. To maTce an isometrical drawing of a pyramid 
standing upon a recessed pedestal. PI. X., Fig. 109. (From a 



86 PROBLEMS INVOLVING NON-ISO METRICAL LINES. 

Model.) Let the scale be |. Assuming C, construct the isometric 
square, CABD, of which each side is 4£ inches. From each of its 
corners lay off on each adjacent side I± inches, giving points, as e 
and/"/ from all of these points lay off, on isometric lines, distances 
of f of an inch, giving points, as g, Jc, and h. From all these points 
now found in the upper surface, through which vertical lines can be 
seen, draw such lines, and make each of them one inch in length, 
and join their lower extremities by lines parallel to the edges of the 
top surface. 

Isometric lines through g, m, &c, give, by their intersections, the 
corners of the base of the pyramid, that being in accordance with 
the construction of this model, and the intersection of AD and CD 
is o, the centre of that base. The height of the pyramid — If inches 
— is laid off at ov. Join v with the corners of the base, and the 
construction will be complete. 

217. To find the shadows on this model. — According to prin- 
ciples enunciated in Division III., the shadow of fh begins at h, 
and will be limited by the line hs, s being the shadow off, and the 
intersection of the ray fs with hs. st is the shadow of/*F. Accord- 
ing to (207), w is the shadow of v, and joining w with p and q, the 
opposite corners of the base, gives the boundary of the shadow on 
the pedestal, ka is the boundary of the shadow of gk on the face 
leu. The heavy lines are as seen in the figure. If this drawing is 
to be shaded, the numerals will indicate the darkness of the tint for 
the several surfaces — 1 being the lightest. 

218. Phob. 8. To construct the isometrical drawing of a wall in 
batter, with counterforts, and the shadows on the wall. PI. X., 
Fig. 110. A wall in batter is a wall whose face is a little inclined 
to a vertical plane through its lower edge, or through any horizon- 
tal line in its face. Counterforts, or buttresses, are projecting parts 
attached to the wall in order to strengthen it. Let the scale be 
one fourth, i.e., let each one of the larger spaces on the scale 
marked 40, on the ivory scale, be taken as an inch. 

Assume C, and make CD = 2 inches and CI= 10 inches in the right 
and left hand isometrical directions. Make DF=6 inches, EF=1|- 
inches, FG=10 inches, and GII=1^ inches, and join H and E, all 
of these being isometrical lines. Next draw EC, then as the top 
of the counterfort is one inch, vertically, below the top of the wall, 
while EC is not a vertical line, make Fc/=one inch, draw de paral- 
lel to FE and ef parallel to EH. Make ef and Ca each one inch, 
at /make the isometric line fg— I of an inch, and at a make ab=^2. 



PROBLEMS INVOLVING NON-ISOMETEICAL LINES. 87 

inches, and draw bg. Make bc=l^ inches, draw ch parallel to bg 
and complete the top of the counterfort. The other counterfort 
is similar in shape and similarly situated, i. e., its furthermost lower 
corner, k, is one inch from I on the line IC, while HI is equal and 
parallel to CE. 

219. To find the shadows of the counterforts on the face of the 
wall. — We have seen — PI. VILT., Fig. 78 — that when a line is per- 
pendicular to a vertical plane, its shadow on that plane is in the 
direction of the projection of the light upon the same plane, and 
from PI. IX., Fig. 102, that the projection, an, of a ray on the left 
hand vertical face of a cube makes on the isometrical drawing an 
angle of 60° with the horizontal line nn' . 

Xow to find the shadow of mn. The front face of the wall — PL 
X., Fig. 110 — not being vertical, drop a perpendicular, fo, from an 
upper back corner of one of the counterforts, upon the edge ba, 
produced, of its base, and through the point o, thus found, draw a 
line parallel to CI. nfop is then a vertical plane, pierced by mn, 
an edge of the further counterfort which casts a shadow; and 
np is the direction of the shadow of mn on this vertical plane. 
The shadow of mn on the plane of the lower base of the wall is of 
course parallel to mn, and^ is one point of this shadow, hence pq 
is the direction of this shadow. Xow n, where mn pierces the 
actual face of the wall, is one point of its shadow on that face, and 
q, where its shadow on the horizontal plane pierces the same face, 
is another point, hence nq is the general direction of the shadow 
of mn on the front of the wall, and the actual extent of this shadow 
is nr, r being where the ray mr pierces the front of the wall. 

From r, the real shadow is cast by the edge, mm, of the counter- 
fort, r, the shadow of m, is one point of this shadow, and s, where 
am produced, meets yn produced, is another (in the shadow pro- 
duced), since s is, by this construction, the point where um, the 
line casting the shadow, pierces the surface receiving the shadow. 
Hence draw srt, and nrt is the complete boundary of the shadow 
sought. The shadow of the hither counterfort is similar, so far as 
it falls on the face of the wall. 

Other methods of constructing this shadow may be devised by 
the student. Let t be found by means of an auxiliary shadow of um 
on the plane of the base of the wall. 

Remark. — In case an object has but few isometrical lines, it is 
most convenient to inscribe it in a right prism, so that as many of 
its edges, as possible, shall lie in the faces of the prism. 



CHAPTER IV. 

PROBLEMS INVOLVING THE CONSTRUCTION AND EQUAL DIVISION OF 
CIRCLES IN ISOMETRICAL DRAWING. 

220. Prob. 9. To make an exact construction of the isometrical 
drawing of a circle. PI. X., Figs. 111-112. This construction is 
only a special application of the general problem requiring the con- 
struction of points in the isometric planes. 

Let PL X., Fig. Ill, be a square in which a circle is circum- 
scribed. The rhombus — Fig. 112 — is the isometrical drawing of the 
same square, CA being equal to CA'. The diameters g'hJ and e'f 
are those which are shown in their real size at gh and ef giving 
g, A, e, and /as four points of the isometrical drawing of the circle. 
In Fig. Ill, draw b'a', from the intersection, b\ of the circle with 
A'D' and parallel to C'A'. As a line equal to b'a\ and a distance 
equal to A! a can be found at each corner of Fig. Ill, lay off each 
way from each corner of Fig. 112, a distance, as A<2, equal to AV, 
and draw a line ab parallel to CA and note the point b, where it 
meets AD. Similarly the points n, o, and r may be found. Having 
now eight points of the ellipse which will be the isometrical draw- 
ing of the circle, and knowing as further guides, that the curve is 
tangent to the circumscribing rhombus at g, h, e and f and perpen- 
dicular to its axes at 5, n, o and r, this ellipse can be sketched in by 
hand, or by an irregular curve. 

221. If, on account of the size of the figure, more points are desira- 
ble they can readily be found. Thus ; on any side of Fig. Ill, take 
a distance as CV and c'd' perpendicular to it, and meeting the circle 
at d'. In Fig. 112, make Cc=CV, and make coequal to c'd' and 
parallel to CD, then will d be the isometrical position of the 
point d'. 

222. Prob. 10. To maTce an approximate construction of the iso- 
metrical draioing of a circle. PI. X., Fig. 113. By trial we shall find 
that an arc, gf having Cfor a centre and C/*for its radius, will very 
nearly pass through n / likewise that an arc eh, with B for a centre, 
will very nearly pass through r. These arcs will be tangents to the 



PIilX. 




PROBLEMS INVOLVING THE ISOMETRICAL DRAWING OF CIRCLES. 89 

sides of the circumscribing square at their middle points, as they 
should be, since C/and Be are perpendicular to these sides at their 
middle points. Now in order that the small arcs, fbh and goe i should 
be both tangent to the former arcs and to the lines of the square at 
g, A, f and e, their centres must be in the radii of the larger arcs, 
hence at their intersections p and q. Arcs having^ and q for cen- 
tres, and ph and qe as radii, will complete a four-centred curve 
which will be a sufficiently near approximation to the isometrical 
ellipse, when the figure is not very large, or when the object for 
which it is drawn does not require it to be very exact. 

223. Prob. 11. To maJce an isometrical drawing of a solid com- 
posed of a short cylinder capped by a hemisphere. PL X., Fig. 114. 
Scale =±. Let this body be placed with its circular base lowermost, 
as shown in the figure. Make ac and bd the height of the cylin- 
drical part=l T 5 2 inches, and draw cd. Now a sphere, however 
looked at, must appear as a sphere, hence take e, the middle point 
of cd, as a centre, and ec as a radius, and describe the semicircle 
ehd, which will complete the figure. 

224. In respect to execution, in general, of the problems of this 
Division, a description of it is not formally distinguished from that 
of their construction, since the figures generally explain themselves 
in this respect. In the present instance, the visible portion of the 
only heavy line required will be the arc anb. As there is no angle 
at the union of the hemisphere with the cylinder — see the preceding 
problem — no full line should be shown there, but a dotted curve 
parallel to the base and passing through c and d, might be added to 
show the precise limits of the cylindrical part. 

225. Again, if it be desired to shade this body, the element, ny^ 
of the cylindrical part, with the curve of shade, pfry, on the sphe- 
rical part, will constitute the darkest line of the shading. The curve 
of shade, pfry, is found approximately as follows. The line ny being 
the foremost element of the cylinder, yy'—ne, is the projection of 
an actual diameter of the hemisphere, mm" and gg% parallel to ST, 
are the radii of small semicircles of the hemisphere, to which projec- 
tions of rays of light may be drawn tangent, and m' and g\ are 
the true positions of their centres — ym' being equal to nm, and 
yg' equal to ng. Drawing arcs of such semicircles, and drawing 
rays, fd and rS, tangent to them, we determine / and r, points of 
the curve of shade on the spherical part of the body, through 
which, with p and y, the curve may be sketched. 

226. Prob. 12. To construct the isometrical circles on the three 



90 PROBLEMS INVOLVING THE ISOMETEICAL DEA.WING OF CIRCLES. 

visible faces of a cube, as seen in an isometrical drawing. PI. 
XL, Fig. 115. This figure needs no minute description here, 
being given to enable the student to become familiar with the 
position of isometrical circles in the three isometrical planes, and 
with the positions of the centres used in the approximate construc- 
tion of those circles. By inspection of the figure, the following 
general principle may be deduced. The centres of the larger arcs 
are always in the obtuse angles of the rhombuses which represent 
the sides of the cube, and the centres of the smaller arcs are at the 
intersection of the radii of the larger arcs with the diagonals 
joining the acute angles of the same rhombuses — i. e. the longer 
diagonals. 

227. Peob. 13. To make the isometrical drawing of a bird house. 
PI. XI., Fig. 116. Assuming C, make CA'=rl6 inches, Qa—Z inches, 
and CB = 9 inches. At a, make ab—\ inch, and ac=z 8 inches. Draw 
next the isometric lines BD and cD. Through b make 5E = 1 6 inches, 
make Ex4.=:3 inches, and EF=r7 inches. Then draw the isometric 
lines DH and FH. Bisect &E at 1ST, make N/^=ll-J- inches, draw 
cf and ¥f make ce=¥h=fg=oue inch, and draw eg and hg. 
Through A and b draw isometric lines which will meet, as at a'. 
On b¥ make bk, Im and n¥ each equal to 3 inches, and let kl and 
mn each be 3'J inches. At I and n draw lines, as Iv, parallel to CB, 
and one inch long, and at their inner extremities erect perpendicu- 
lars, each 3^ inches long. Also at k, I, m and n, draw vertical iso- 
metrical lines, as kt, 3% inches long. The rectangular openings thus 
formed are to be completed with semicircles whose real radius is 
If inches, hence produce the lines, as kt — on both windows — mak- 
ing lines, as kG, 5£ inches long, and join their upper extremities as 
at GI. The horizontal lines, as ts, give a centre, as s, for a larger 
arc, as tu. The intersection of Go with Iz — see the same letters on 
Fig. 115 — gives the centre, jp, of the small arc, uo. The same ope- 
rations on both openings make their front edges complete. Make 
oa and pr parallel, and each, one inch long, and r will be the centre 
of a small arc from a which forms the visible part of the inner edge 
of the window. Suppose the corners of the platform to be rounded 
by quadrants whose real radius is 1+ inches. The lines a'b and 
bk each being 3 inches, k is the centre for the arc which repre- 
sents the isometric drawing of this quadrant, whose real centre 
on the object, is indicated on the drawing at y. So, near A, w is 
the centre used in drawing an arc, which represents a quadrant 
whose centre is x. — See the same letters on Fig. 115. 



PROBLEMS INVOLVING THE ISOMETRICAL DRAWING OF CIRCLES. 91 

Of the Isometrical Drawing of Circles which are divided in 
Equal Parts. 

228. Prob. 14. PL XL, Fig. 117. First method. — If the semi- 
ellipse, ADB, be revolved up into a vertical position about AB as an 
axis, it will appear as a semicircle AD'B of winch ADB is the iso- 
metrical projection. Since AB, the axis, is parallel to the vertical 
plane, the arc in which any point, as D, revolves, is in a plane per- 
pendicular to the vertical plane, and is therefore projected in a 
straight line DD'. Hence to divide the semi-ellipse ADB into parts 
corresjDonding to the parts of the circle which it represents, divide 
AD'B into the required number of equal parts, and through the 
points thus found, draw lines parallel to D'D, and they will divide 
ADB in the manner required. The opposite half of the curve can 
of course be divided in a similar manner. 

229. Second method. — CE is the true diameter of the circle of 
which ADB is the isometrical drawing. Let it also represent the 
side of the square in which the original circle to be drawn is inscribed. 
The centre of this circle is in the centre* of the square, hence at O, 
found by making eO equal to half of CE, and perpendicular to that 
line at its middle point e. 

With O as a centre, draw a quarter circle, limited by CO and EO, 
and divide it into the required number of parts. Through the points 
of division, draw radii and produce them till they meet CE. CE, 
considered as the side of the isometrical drawing of the square, is 
the drawing of the original side CE of the square itself with all 
its points 1, 2, 6, 7, &c, and O' is the isometrical posi- 
tion of O. Hence connect the points on CE with the point O' and 
the lines thus made will divide the quadrant BC in the manner 
required. 

Applications of the preceding Problem. 

230. Prob. 15. To make an isometric drawing of a segment oj 
an Ionic Column. PI. XL, Fig. 118. Let aD be a side of the cir- 
cumscribing prism of the column. By the second method of Prob. 
14, find O', the centre of a section of the column, and with O' as a 
centre, draw any arc, as a'q'. The curved recesses in the surface 
of a column are called flutes, or the column is said to be fluted. In 
an Ionic, and in some other styles of columns, the flutings are semi- 
circular with narrow flat, or strictly, cylindrical surfaces, as ee"p, 
between them. Llence, in Fig. 118, assume a'b', equal to q'v\ as 
half of a space between two flutes, divide b'v' into four equal parts, 
and make the points of division central points of the spaces as f'e' 



92 PROBLEMS INVOLVING THE ISOMETRICAL DRAWING OF CIRCLES. 

between the flutes. Let the flutes be drawn with points, as c' as 
centres and touching the points as b'd' ; then draw an arc tangent, 
as at r, to the flutes. To proceed now with the isometrical draw- 
ing, draw, in the usual way, the isometrical drawing of the outer 
circumferences of the column, tangent to aD and b r "¥ — assuming 
DF for the thickness of the segment. Now a'q' being any arc, and 
not one tangent to aD so as to represent the true size of a quadrant 
of the outer circumference, the true radius of the circle tangent to 
the inner points of all the flutes will be a fourth proportional, 0'y\ 
to O'f, Oi ( = 0'y), and O's. On Oi, lay off OY=Oy', draw IJ to 
find a centre I, and similarly find the other centres of the larger arcs 
of the inner ellipse. The points n, h and n', ti are the centres of the 
small arcs (222) for the two bases. Having gone thus far, produce 
O'b', O V, &c. to aD ; at b, c, &g., erect vertical lines, bb'", cc"\ 
&c, then from b, c, &c. draw lines to O, and note their intersec- 
tions, b", c", &g. with the curves of the lower base ; and from 
b'", c'", &g. draw lines to O" and note their intersections, b' ! ", c"", 
&g. with the ellipses of the upper base. This process gives three 
points for each flute by whi<5h they can be accurately sketched in, 
remembering that they are tangent to the inner dotted ellipses, as 
at c"", o'", &c. and to the radii, as e"0" — at e" , Parts beyond 
FO" are projected over from the parts this side, thus drawn. 

231. Prob. 16. To construct the isometrical drawing of a seg- 
ment of a Doric Column. PL XI., Fig. 119. The flutes of a Doric 
column are shallow and have no flat space between them. Adopt- 
ing the first method of Prob. 14, let the centre, A, of the plan be 
in the vertical axis, GA', of the elevation, produced. Let Ac and 
Ab be the outer and inner radii containing points of the flutes. 
Make Ad—^ of Ac, for the radius of the circle which shall contain 
the centres of the flute arcs. Let there be four flutes in the qua- 
drant, shown in the plan. Their centres will be at h, &c., where 
radii Ag, &c, bisecting the flutes, meet the outermost arc. In pro- 
ceeding to construct the isometrical drawing, project b and c, at b* 
and c' on the axis A'd'. Now, owing to the variation at b and c' 
between the true and the approximate ellipse, we cannot make use 
of the latter, if we retain b' and c' in their proper places, as projected 
from b and c, hence through b' and c' draw isometric lines which 
locate the points N' and Q' (the points are between these letters) 
which are the true positions of N" and Q respectively. Correspond- 
ing points, between N'" and v, are similarly found. By an irregular 
curve the semi-ellipses vb'Q,' and N"VN' can be quite accurately 



PROBLEMS INVOLVING THE ISOMETRICAL DRAWING OF CIRCLES. 93 

drawn. Next, project upon these curves the points u, e, &c, r, g, 
&c. of the flutes — as at u\ e\ &c, r'\ &c, and with an irregular 
curve draw the curves through these points, tangent to the inner 
semi-ellipse. The corresponding curves of the lower base are found 
by drawing lines r'r'\ u'u'\ &c. through the points of tangency ; 
r\ ~k\ &c, and through u\ &c, and all equal to FD, the thickness 
of the segment. 

The curves above the axis A'd' are projected across from those 
already made below it. 

Special Examples . 

232. Prob. IV. To draw a cube or other parallelopipedical body 
so as to show its under side. PI. XL, Fig. 120. By reflection, it 
becomes evident that it is the relative direction of the lines of 
the drawing among themselves, that make it an isometrical draw- 
ing. Hence in the figure, where all the lines are isometric lines, the 
whole is an isometric drawing, now that the solid angle C is nearest 
us, as much as if the angle A (lettered C on previous figures) were 
nearest us. 

233. Remark. By a curious exercise of the will, we can make 
Fig. 120 appear as an interior view, showing a floor CFED, and 
two walls; or, in Fig. 115 and others, we can picture to ourselves 
an interior showing a ceiling GL& and two walls. This is probably 
because — 1st. All drawings being of themselves only jDlane figures, 
we educate the eye to see in them, what the mind chooses to conceive 
of, as having three dimensions. 2nd. When, as in isometrical draw- 
ing, the drawing in itself as a plane figure, is the same for an interior 
as for an exterior view of any given magnitude, the eye sees in it 
whichever of these two the mind chooses to imagine. 

234. Prob. 18. To construct isometrical drawings of oblique 
sections of a right cylinder with a ciradar base. PI. XL, Fig. 
121. This construction is easily made from a given circle as a base 
of the cylinder, that base being in an isometric plane. The circle in 
the plane AGEF is such a circle. Let A'G'E'F' be a plane inclined 
to AGEF but perpendicular, as the latter is, to the planes GB and 
DF, and let A'G'E'F' be a plane inclined to all the sides of the 
prism AGE— D. 

Lines, as aa'a\ &c, being in the faces of the prism and parallel 
to their edges, meet the intersections, F'E'— F'E", &c. of the 
oblique planes at points a\ a\ &c, which are points of oblique 
sections of a cylinder inscribed in the prism AGE — D, and wnose 
base is acbdu. 



94 PROBLEMS INVOLVING THE ISOMETRIC AL DRAWING OF CIRCLES. 

So, points, as c, have the corresponding points c'c", &c. in the 
diagonals A'E', A"E" of the planes in which those points are found. 

To find points, as t\ t" , &c. corresponding to t in the base, draw 
any line, as yd, through t, and find the corresponding lines, as y'd r 
and y"d". Their intersections with the diagonals G'F' and G"F" 
will give the points t\ t\ &c. Having thus found eight points of 
each oblique section of the given inscribed cylinder whose base is 
abcd-u, and remembering that each of these sections is tangent to the 
sides of its circumscribing polygon (considering the lines y'd\&c), 
the curves a', b/ c\ t\ and a", b", c", t" are readily sketched in. 

235. Remarks, a. As before stated, it is the relative direction, 
among themselves, of the lines of an isometrical drawing, that deter- 
mine it as an isometrical drawing, hence PI. XI., Fig. 121, is an 
isometrical drawing, though its lines are not situated with refer- 
ence to the edges of the plate as the similar lines of previous 
figures have been. If the portion of the plate containing this figure 
were cut out so as to make the edges of the fragment, so cut out, 
parallel and perpendicular to GE, the figure would appear like 
the previous isometrical drawings. 

b. The problem just solved must not be confounded with one 
which should seek to find the isometric projection of a curve 
which in space is a circle on the plane G'E' — A', for the curve 
a'b'c'd't' is not a circle, in space. 

236. Prob. 19. To solve the problem just enunciated. PI. XL, 
Figs. 121-122. eV— Fig. 122— is a plan of the section rA'F' in 
which — it being a square — a circle can be inscribed. e"r is there- 
fore the plan of the circle also. Making rG — Fig. 122 — equal to 
rG' — Fig. 121, and drawing e"G, we have the plan of the section 
G'E' — A', and making o"p\ Fig. 122, equal to e'V, we have the plan 
of a circle in the section G'E' — A'. Now draw o'x and p'e' — Fig. 
122 — make A'e and GV" and e'"p and eo — Fig. 121 — equal to e r/ x, 
Ge', e'p' and xo"— 'Fig. 122 — drawjt?Y and oU; and u'TJ and #'Y 
to intersect them, and we shall have U and Y as the isometric 
positions in the plane G'E' — A' of the points o' and p' which, 
considered as points on the circle, are evidently enough extre- 
mities of its horizontal diameter, at which points, the circle is tan- 
gent to the vertical lines whose isometric positions in the plane 
G'E' — A' are pY and oU. T and a' are other points. 

The finding of intermediate points, which is not difficult, is left as 
an exercise for the student. 



CHAPTER Y. 



CABIXET PEOJECTIOXS. 



1. There is a kind of projection, examples of which, in the draw- 
ing of details, etc., are oftener seen in French works than isometri- 
cal projection (an English invention) is. It has been variously 
named, "Military," "Cavalier," or "Mechanical" Perspective. 
We propose to call it "Cabinet Projection," it being especially 
applicable to objects no larger than those of cabinet work, and 
being actually used in representing such work. 

2. This new projection differs from isometrical, chiefly in show- 
ing one of a number of pairs of parallel faces, of a cube, for exam- 
ple, in their true form as well as dimensions. 

Thus ; Fig. 1 is the isometrical drawing of a cube, and Fio-. 2 is 





Vis. 2. 



a cabinet projection of the same cube ; all the edges being of the 
same length in both figures. Hence we see, as stated, that in the 
latter figure, the faces DEFG, and ABCH, and by consequence 
every line in them, are shown in their true form, as well as size ; 
which is not true of isometrical drawing. 

3. The parallel oblique lines, FA, etc., are here made equal to 
ED, an edge of the cube, and at an angle, AFG, of 45° with DE. 



Qg CABINET PROJECTIONS. 

In other similar cases, edges, as FA, would still be shown in their 
true length, as in isometrical drawing, whenever they were in 
realitv perpendicular to the face, as DEFG. 

Other values than 45° for the angle AFG, will be discussed 

presently. . 

4. Another advantage of cabinet projection, already apparent, 
is, that the remote corner, H, which, in the isometrical drawing of 
a 'cube, coincides with the foremost corner, G, is seen separately 
in the cabinet projection. 

Also, of the four body diagonals of the cube, one, GH, appears 
as a point only, in isometrical projection, and the other three, as 
FC, are all partly confounded with the projections, as FG, of 
edges. But in the cabinet projection, all these diagonals show as 
lines, and, except BE, separately from the edges of the cube. 

5 Having thus shown something of what cabinet projection is, 
let us proceed to learn how it is formed. Simply remembering 
that a line, as AB, Fig. 3, which makes angles of 45° with lines, as 



Fi£. 3. 



AC and BC, which are perpendicular to each other, cuts off equal 
distances, CA and CB, from their intersection, C ; it is evident 
that the eye is, in reference to Fig. 2, at an infinite distance from 
the paper, and above and to the right of the object, in a plane per- 
pendicular to the paper in the line EB, and that the direction of 
vision makes an angle o/45° with the paper, taken as the plane of 

projection. 

Thus EH^ED, and is the cabinet projection of the edge which 
is really perpendicular to the paper at E, and has the point E for 

its vertical projection. ,.,.-«. \ J 

6 We have thus a system of projections which differ from all 

those already noticed, in that the projecting lines are oblique to 



PL.X 



/2.?-» p a 




The Major & Knavp Ene.Mfa.ALitli.fn 4+9 Broadway NT. 




II- M.»j»i.vKii.i,,].h...,\ii, ■. I,rl,r. 1 .|l!l!l l i-„K.., 1 .', \ 



CABINET PROJECTIONS. 



97 



the plane of projection. Such are true oblique projections ; though 
bodies placed obliquely to the planes of projection are sometimes 
said to be shown in oblique projection. But in truth they are only 
seen obliquely, while they are projected perpendicularly, that is by 
projecting lines which are perpendicular to the planes of projec- 
tion. 

The two systems of projection, in which the eye is at an infinite 
distance, are, therefore, perpendicular projection, in which the 
projecting lines are perpendicular to the planes of projection ; and 
oblique projection, in which they are oblique to those planes. 

Isometrical projection is thus seen to be a variety of true perpen- 
dicular projection, in which the object is, however, seen obliquely, 
and in which each plane side of a solid right angle is, indeed, 
viewed with equal obliquity. 

Cabinet projection is a species of oblique projection. 

7. Lines making angles of 45° with the vertical plane, may make 
various angles with the horizontal plane ; we therefore now pro- 
ceed, as was proposed, to compare the various directions which the 
line FA, Fig. 2, may have ; considering only 45°, 30°, and 60° as 
the values of AFG=CDK. 



/I 


/ 




c 


Hi ? 




r 


7 \ 


bl 


/ j 


i* 


/ 


-i 


/ 


.** 


/ ! 


''•' ' 1 • !■"' ' 1 ■ ' 


/. J 




Kg. 4 



Figs. 4, 5, and 6, are three different cabinet projections of the 
same cube. In Fig. 4, DK=KC=iv/27when CD=DE=1. That 
is, KC=half the diagonal of a square whose side=CD. This 
enables us to draw HC and AB indefinitely, to limit the oblique 
lines, DC, etc., when their length is given. 

In Fig. 5, CDK=30°,and CK=sin 30° to the radius DC, =i DC; 
knowing which, HC and AB can be drawn as before, to limit FA, 
DC, etc. 

1 



98 



CABINET PROJECTIONS. 



In Fig. 6, CDK = 60°. Accordingly DK=iDC, from which, 
having found K, the perpendicular KC can be drawn to limit DC. 
Or, as before, CK=i v% DC being ==i. That is, CK = AEC, since 
CDE = 120 . And for a square prism of any length, KC= half of 
the diagonal joining alternate vertices of a regular hexagon whose 
side equals the edge of the prism, lying in the direction of DC, and 
whose length is supposed to be given. 



F 


A 






C 


Hj 








L 


[ 


) 



Fig. 6. 

With these illustrations, the student might proceed to investi- 
gate other relations between the parts of these, and still other 
cabinet projections. But the above may suffice for now. 

We observe that, in Figs 5 and 6, none of the body diagonals 
are confounded with the edges ; and that each of the tnree forms 
maybe preferable for certain objects. 

8. Points not on the axes EF, EH, and ED, or on parallels to 
them, are found by co-ordinates, as in isometrical drawing. Thus, 
if ED, Fig. 4, be 4 inches, and if we make Ea=2 inches, ab paral- 
lel to EH, = 2 \ inches, and be, parallel to EF, = 1^ inches, then c is 
the cabinet projection of a point, 2 inches from the face FH ; *>^ 
inches from the face FEG, and 1^ inches above the base EDC. 
This principle will enable the student to reconstruct any of the 
preceding isometrical examples of straight-edged objects, in cabi- 
net projection. 

9. It only now remains to explain the cabinet projections of cir- 
cles. Let Fig. 7 be the cabinet projection of a cube, with circles 
inscribed in its three visible faces. One of these circles, abed, will 
appear as a circle, and so would the invisible one on the parallel 
rear face. 

For the ellipse in BCDG, draw the diagonals, BD and CG, of 



CABINET PROJECTIONS. 



99 



that face. Then in the cube itself, horizontal lines joining corre- 
sponding points in the circles abed, and hpfu, are parallel to the 
diagonal EC. Hence tit, ef, mp, and gh determine the points w, f, p, 




Fig. 1. 



and A, by their intersections with the diagonals BD and CG. The 
middle points of the sides of the face BCDG, are also points of the 
ellipse, and are its points of contact with those sides. The ellipse 
also has tangents at h andjf, parallel to BD, and at u and p, paral- 
lel to CG. Hence, having eight points, all of which are points of 
contact of known tangents, the ellipse can be accurately sketched. 

10. The ellipse in the upper face could be found in the same 
manner. But an approximate construction by circular arcs has 
been shown, to test its accuracy and appearance, as compared with 
the approximate isometrical ellipse. The ellipse being tangent afc 
L and K, perpendiculars to FA and BA, at those points, will in- 
tersect at M, the centre of an arc tangent to FA and BA at those 
points. Then a~N perpendicular to FG at a, and equal to MK, 
gives N, the centre of the arc an. As the remaining arcs must be 
tangent to those just drawn, their centres, r and s, must be the 
intersections of the radii of the large arcs, with the transverse axis, 
FB, of the ellipse. 

The true extremities of the transverse axis are found by drawing 



100 CABINET PROJECTIONS. 

pq parallel to AC, and a parallel to it from u. The error qq' at 
each end of the transverse axis, is thus seen to be considerable. 
Also the greater difference between the radii, than occurs in mak- 
ing the isometrical ellipse, occasions a harsh change of curvature at 
K, "N, etc. ; so that the approximate construction of the cabinet 
ellipse is of very little value. 

11. It is found on trial, that the centre M falls both on BD and 
EC, so that neither KM nor LM really need be drawn. The reason 
of this property, which so simplifies the construction, is evident. 
For, BA=AF=FE are in position as three sides of a regular octa- 
gon, so that the perpendiculars, as KM, from the middle points of 
those sides, will meet at the same points with BD, AGM, EC, etc., 
which are obviously the bisecting lines of the angles of the octa- 
gon, viz. at the centre, M, of the octagon. 

By varying the angle GFA, as in the previous figures, the stu- 
dent may discover similar coincidences, which he can explain for 
himself. 

12. Finally, it is to be noticed, that the pictorial diagrams of PI. 
I., Figs. 1, 2, 3, 5, etc., which are so effective a substitute for 
actual models, to most eyes, are merely cabinet projections of 
models themselves. 

The student having well understood this chapter, as well as the 
practice of isometrical drawing, will be prepared to make the 
cabinet projection of any object, without further illustration. 



DIVISION FIFTH. 

ELEMENTARY STRUCTURAL DRAWING. 



237. Note. The objects of this Division are, to acquaint the 
student with a few things respecting the drawing of whole structures 
which are not met with in the drawing of mere details; to serve as 
a sort of review of practice in certain processes of execution ; and 
to afford illustrations of parts of structures whose names have yet 
to be defined. Proceeding with the same order as regards material 
that was observed in Division Third, we have : — 

CHAPTER I. 

STONE STRUCTURES. 

238. Example 1°. A "brick segmental Arch. PI. XII, Fig. 123. 
Description of the structure. — A segmental arch is one whose 

curved edges, as «Cc, are less than semicircles. A brick segmental 
arch is usually built with the widths of the bricks placed radially, 
since, as the bricks are rectangular, the mortar is disposed between 
them in a wedge form in order that each brick with the mortar 
attached may act as a wedge ; while if the length of the bricks be 
radial, the mortar spaces will be inconveniently wide at their outer 
ends, unless the arch be a very wide one, or unless it have a very 
large radius. 

The permanent supports of the arch, as wPT, are called abut- 
ments, and the radial surface, as nab, against which the arch rests, 
is called a skew-back. 

The temporary supports of an arch while it is being built are 
called centres or centrings, and vary from a mere curved frame 
made of pieces of board — as used in case of a small drain or round 



102 STONE STRUCTURES. 

topped window — to a heavy and complicated framing, as used for 
the temporary support of heavy stone bridges. 

Note. The general designing of these massive centrings may 
call for as much of scientific engineering knowledge, and their details 
and management may call for as much practical engineering skill, 
as does the construction of the permanent works to which these 
centrings are auxiliary. In short, the detailed design and manage- 
ment of auxiliary constructions, in general, is no unimportant depart- 
ment of engineering study. 

The span is the distance, as ac, between the points of support, on 
the under surface of the arch. The stones over the arch and abut- 
ment, form the spandril, or backing, Qc?P. 

239. Graphical construction. — Let the scale be one of four feet 
to the inch =48 inches to one inch = 5*5-. Draw RT to represent 
the horizontal surface on which the arch rests. Let the radius of 
the inner curve of the arch be 1 feet, the height of the line ac from 
the ground 2 feet 8 inches, and the span 7 feet. Then at some 
point of the ground line, draw a vertical line, OC, for a centre 
line ; then draw the abutments at equal distances on each side of 
the centre line, and 6 feet 8 inches apart. Let them be 2 feet 6 
inches wide. 

Since the span and radius have been made equal, Ob and Od may 
be drawn, in this example, with the 60° triangle. Drawing these 
lines, and making Oa~7 feet, make ab = one foot, draw the two 
curves at the end of the arch, and make b and d points in the top 
surfaces of the abutments. 

To locate the bricks, since the thickness of the mortar between 
the bricks, at the inner curve of the arch, would be very slight, lay 
off two inches on the arc aCc an exact number of times. The dis- 
tance taken in the compasses as two inches, may be so adapted as 
to be contained an exact number of times in aCc, since the thick- 
ness of the mortar has been neglected, but would in practice be so 
adjusted, as to allow an exact number of whole bricks in each 
course. 

The arch being a foot thick, there will be three rows of bricks 
seen in its front. Draw therefore two arcs, dividing ab and cd into 
three spaces of four inches each, and repeat the process of division 
on both of them. 

Having all the above-named divisions complete, fasten a fine 
needle vertically at O, and, keeping the edge of the ruler against it, 
to keep that edge on the centre without difficulty, draw the lines 
which represent the joints in each of the three courses of brick. 



ST0XE STRUCTURES. 103 

240. Ex. 2°. A semi-eylindrical Culvert, having vertical 
quarter-cylindrical Wing Walls, truncated obliquely. PI. 
XII., Fig. 124. 

Description of the structure. — A culvert is an arched passage, 
often flat bottomed, constructed for the purpose of carrying water 
under a canal or other thoroughfare. Wing walls are curved con- 
tinuations of the vertical flat wall in which the end of the arch is 
seen. Their use is to support the embankment through which the 
culvert is made to pass, and to prevent loose materials from the 
embankment from working their way or being washed into the cul- 
vert. Partly, perhaps, for appearance's sake, the slope of the plane 
which truncates the flat arch-wall, called the spandril wall, and the 
wing walls, is parallel to the slope of the embankment. The wing 
walls are often terminated by rectangular flat-topped posts — "piers" 
or "buttresses," AA', and the tops, both of these piers and of the 
walls, are covered with thin stones, abed — a"b"c"d", broader than 
the wall is thick, and collectively called the coping. 

Since the parts of stone structures are not usually firmly bound 
or framed together, each course cannot be regarded as one solid 
piece, but rather each stone, in case, for instance, of the lowermost 
course, rests directly on the ground independently of other stones 
of the same course, hence if the ground were softer in some spots, 
under such a course, than in others, the stone resting on that spot 
would settle more than others, causing, in time, a general disloca- 
tion of the structure. Hence it is important to have what are called 
continuous bearings, that is, virtually, a single solid piece of some 
material on which several stones may rest, and placed between 
the lowest course and the ground. 

Timbers buried away from the air are nearly imperishable ; hence, 
timbers laid upon the ground, if that be firm, and covered with a 
double floor of plank, form a good foundation for stone structures; 
and in the case, of a culvert, if such a flooring is made continuous 
over the whole space covered by the arch, it will prevent the flow- 
ing water from washing out the earth under the sides of the arch. 

When the wing walls and spandril are built in courses of uni- 
form thickness, the arrangement of the stones forming the arch, 
so as to bond neatly with those of the walls, offers some difficul- 
ties, as several things are to be harmonized. Thus, the arch stones 
must be of equal thickness, at least all except the top one, and then, 
there must be but little difference between the widths of the top, 
or key stone, and the other stones ; the stones must not be dis- 
proportionately thin or \evy wide, they should have no re-entrant 



104 STONE STRUCTURES. 

angles, or very acute angles, and there must not be any great 
extent of unbroken joint. 

241. Graphical construction. — Let the scale be that of five feet 
to an inch = 60 inches to an inch=^_. 

a. Draw a centre line, BB', for the plan. 

b. Supposing the radius of the outer surface, or back, of the arch 
to be 5j feet, draw CC parallel to BB' and 5^ feet from it. 

c. Draw BE, and on C'C produced, make EC = 9 feet 8 inches, 
CD, the thickness of the face wall of the arch=2 feet 4 inches, and 
the radius, oD, of the face of the wing wall=4 feet. 

d. With o as a centre, draw the quadrants CG, and DF, and with 
a radius of 3 feet 8 inches, draw the arc cA, the plan of the inner 
edge of the coping. Also draw at D and C, lines perpendicular 
to BB' to represent the face wall of the arch. 

e. At G, draw GA towards o, and =3 feet, for the length of 
the cap stone of the buttress, AA', and make its width =2 feet 
10 inches, tangent to CG at G. The top of this cap stone, being a 
flat quadrangular pyramid, draw diagonals through G and A, to 
represent its slanting edges. 

f. Supposing the arch to be 1^ feet thick, make C'H=l£ feet, 
and at C and H, draw the irregular curved lines of the broken end 
of the arch, and the broken line near the centre line, also a fragment 
of the straight part of the coping. 

g. Let the horizontal course on which the arch rests, be 2 feet 9 
inches wide, i. e., make He=3 inches, and CVz=l foot ; and let the 
planking project 3 inches beyond the said course, making er=3 feet. 
Through e, n and r, draw lines parallel to BB' and extending a little 
to the right of C'H. 

A. Proceeding to represent the parts of the arch substantially in 
the order of their distance from the eye, as seen in a plan view, a 
portion of the planking may next be represented. The pairs of 
broken edges, and the position of the joints, show that there are 
two layers of plank and that they break joints. 

i. Under these planks, appear the foundation timbers, which 
being laid transversely, and being one foot wide and one foot apart, 
are represented by parallels one foot apart, and perpendicular to 
BB'. Let the planking project 4 inches beyond the left hand tim- 
ber. Observe that two timbers touch each other under the arch 
front. 

j. The general arrangement of stones in the curved courses of the 
wing wall, in order that they may break joints, is, to have three and 
four stones, respectively, in the consecutive courses. To indicate 



STONE STKTTCTURES. 105 

this arrangement in the plan, 7iG,fb, gd and DC will represent the 
joints of alternate courses, and the lines km, &c. midway between 
the former, will represent the joints of the remaining intermediate 
courses. 

This completes a partial and dissected plan which shows more of 
the construction than would a plan view of the finished culvert, and 
as much, as if the parts on both sides of the centre line were shown. 
In fact, in drawings which are strictly working drawings, each pro- 
jection should show as much as possible in regard to each distinct 
part of the object represented. 

242. Passing to the side elevation, which is a sectional one, show- 
ing parts in and beyond a vertical plane through the axis of the 
arch, we have : — 

a. The foundation timbers, as m'q, &c, projected up from the 
plan ; or, one of them being so projected, the others may be con- 
structed, independently of the plan, by the given measurements. 

b. The double course of planking op, appears next with an occa- 
sional vertical joint, showing where a plank ends. 

c. The buttress, A, and its cap stone Y, are projected up from 
the plan, and made 6 feet high, from the planking to G\ 

d. From G' and h', the slanting top of the wing walls are shown, 
as having a slope of 1% to 1 — i. e. h'h"=% ti'u — and the vertical lines 
at C, D' and D" are projected up from C, D and D"\ 

The remaining lines of the side elevation are best projected back 
from the end elevation, when that shall have been drawn. 

243. In the end or front elevation, we have : — 

a. At m"m'", a side view of one of the foundation timbers, 
broken at m'", so as to show other timbers behind it. 

b. The planking o'o" in this view, shows the ends of the planks 
in both layers — breaking joints. 

c. No' = ~Bo f ", taken from the plan ; and in general, all the hori- 
zontal distances on this elevation, are taken from the plan, on lines 
perpendicular to BB\ 

d. The vertical sides of the buttress, A', are thus found. The 
heights of its parts are projected over from the side elevation. 

e. The thickness of the foundation course, ts= 1|- feet, and tr'—en, 
on the plan. 

f. The centre, O, of the face of the arch, is in the line r't pro- 
duced. The radius of the inner curve (intrados) of the arch is 4 feet 
and of the cylindrical back, behind the face wall, 5\ feet — shown by 
a dotted arc. In representing the stones forming the arch, it is to 
be remembered that they must be equal, except the " key stone," 



106 STONE STRUCTURES. 

<7, which may be a little thicker than the others; they must also be 
of agreeable proportions, free from very acute angles, or from re- 
entrant obtuse angles ; and must interfere as little as possible with 
the bond of the regular horizontal courses of the wing walls. There 
must also be an odd number of stones (ring stones) in the front of 
the arch. 

On both elevations, draw the horizontal lines representing the 
wing wall courses as one foot in thickness, and divide the inner 
curve of the arch into 15 equal parts. Draw radial lines through 
the points of division. Their intersections with the horizontal lines 
are managed according to the principles just laid down. 

g. The points, as Jc and/, in the plan, are then projected into the 
alternate courses of the side elevation, and into the line, Bo'", of 
the plan. 

From the latter line, the several distances, o'"b"\ &c, from <?'", 
thus found, are transferred to the line o'N, as at o'b"", &c, and at 
these points the vertical joints of the front elevation are drawn in 
their proper position, as being the same actual joints, shown by the 
vertical lines of the side elevation. In the stones immediately under 
the coping, there must generally be some irregularity, in order to 
avoid triangular stones, or stones of inappropriate size. 

A. To construct the front elevation of the coping. All points, as 
a, a', a", in either the front or back, or upper or lower edges of the 
coping, are found in the same way, and as follows : 

a" is in a horizontal line through a r and in a line a" a"", whose 
distance from o' equals the distance o'"a'" on the plan. Construct- 
ing other points similarly, the edges of the coping may be drawn 
with an "irregular curve." 

The horizontal portion of the coping, over the arch, is projected 
over from C and from the two ends of the vertical line at D'. 

JExecution. — In respect to this, the drawing explains itself. 



CHAPTER n. 

WOODEN STRUCTURES. 

244. Ex. 3°. Elevation of a " King Post Truss." 

Mechanical construction, <£c. — A Truss is an assemblage of pieces 
so fastened together as to be virtually a single piece, and therefore 
exerting only a vertical force, due to its weight, upon the support- 
ing walls. 

In PL XII., Fig. 125, A is a tie beam/ B is a principal y C is a 
rafter y D is the king post y E is a strut y F is a wall plate y G is a 
purlin — running parallel to the ridge of the roof, from truss to truss, 
and supporting the rafters. H is the ridge pole y W is the wall^ 
and ab is a strap by which the tie beam is suspended from the king- 
post. 

245. Graphical construction. — In the figure, only half of the truss 
is shown, but the directions apply to the drawing of the whole. In 
these directions an accent, thus ' , indicates feet, and two accents, 
% inches. 

a. Draw the vertical centre line bJ). 

b. Draw the upper and lower edges of the tie beam, one foot 
apart, and 12' in length, on each side of the vertical line. 

c. On the centre line, lay off from the top of the tie beam, 5' — 6" 
to locate the intersection of the tops of the principals ; and on the 
top of the tie beam, lay off 11' on each side, to locate the intersec- 
tion of the upper faces of the principals with the top of the tie beam. 

d. Draw the line joining the two points just found, and on any 
perpendicular to it, zsfg, lay off its depth = 8*, and draw its lower 
edge parallel to the upper edge. Make the shoulder at o = 3" and 
parallel to/*/. 

e. From the top of the beam, draw short indefinite lines, c, 6* 
each side of the centre line, and note the points, as e, where they 
would meet the upper sides of the principals. 

f. Draw vertical lines on each side of the centre line and 4" 
from it. 

g. From the points, as e, draw hues parallel to fg till they inter- 
sect the last named vertical lines. 



108 WOODEN STRUCTURES. 

h. Make ns = 5' — 9". Make the short vertical distance at c=4", 
draw so, and make the upper side of the strut parallel to so, and 4" 
from it. Note the intersection of this parallel with the line to the 
left of D, and connect this point with the upper end of c, to com- 
plete the strut. 

i. Draw the edges of the rafter, parallel to those of the principal, 
4" apart, and leaving 4" between the rafter and the principal. At 
o, draw a vertical line till it meets the lower edge of C, and from 
this intersection draw a horizontal line till it meets the upper edge 
of C ; which gives proper dimensions to the wall plate. 

j. From the intersections of the upper edges of the rafters, lay 
off downwards on the centre line 12", and make the ridge pole, 
thus located, 3" wide. 

k. In the middle of the upper edge of the principal, place the 
purlin 4" x 6", and setting 2" into the principal. 

I. Let the strap, ab, be 2" wide, and 2' — 6" long from the bottom 
of the tie beam. Let it be spiked to the king post and tie beam, 
and let it be half an inch thick, as shown below the beam. W, the 
supporting wall, is made at pleasure. 

Execution. — This mainly explains itself. As working drawings 
usually have the dimensions figured upon them, let the dimensions 
be recorded in small hair line figures, between arrow heads which 
denote what points the measurements refer to. 

246. Ex. 4°. A "Queen Post Truss" Bridge. PI. XIII. 

Fig. 126. 

Mechanical construction. — This is a bridge of 33 feet span, over 
a canal 20' — 6" wide between its banks at top, and 20' — 2" at the 
water line. It rests on stone abutments, R and P, one of which is 
represented as resting on a plank and timber foundation, the other 
on " piles." 

A is the tie beam ; B, B' the queen posts ; C, C the principals ; 
D the collar beam, or straining sill ; R, P, the abutments ; eQt the 
pavement of the tow path ; iK. the stone side walls of the canal; 
TT the opposite timber wall, held by timbers ITU', N, dovetailed 
into the wall timbers ; E, S, the piles, iron shod at bottom. These 
are the principal parts. 

247. Graphical construction. — Let the scale be one of five feet 
to the inch. 

a. All parts of the truss are laid off on, or from, the centre line 
AD. A is 14" deep; the dimensions of BB' are 12" x 6', except at 
top, where they are 10" wide for a vertical space of 16". C and D 



WOODEN STRUCTURES. 109 

are each 10" deep. BB' are 10' apart, and the feet of C and C, 12" 
from the ends of the tie beam, which is 36' long. D is 6" below 
the top of the queen posts, rr are inch rods with five inch washers, 
£" thick, and nuts 2f xl". lib' is a \ n bolt; with washer 4"x£" 
and nuts, 2"xl"; and perpendicular to the joint, ad. 

b. From each end of the tie beam, lay off 1' — 9" each way for the 
width of the abutments, at the top. Make the right hand abutment 
rectangular in section and 11' high, of rectangular stones in irregu- 
lar bond (76). Let the left hand abutment have a batter of 1" in 1' 
on the side towards the canal, and let it be eleven feet high, in 
eleven equal courses. 

c. Make et, the width of the paved tow path = '7 / — 6", with a rise 
in the centre, at Q, of 6". 

d. The side wall is of rubble, 4' thick at bottom, and extending 
18" below the water, with a batter of I" in 1', and having its upper 
edge formed of a timber 12" square. 

e. The right hand abutment rests on a double course of three-inch 
planks, qq\ 5^' broad, and resting on four rows of 10" piles, ES. 
S is the sheet iron conical shoe at the lower end of one of these 
piles, the dots at the upper end of which represent nails which 
fasten it to the pile. 

f. TT is a timber wall having a batter of 1" to 1', and held in 
place by timbers, ITU', N, dovetailed into it at its horizontal joints, 
in various places. 

g. The water line is 2' below Ttf, and the water is 4|- feet deep. 
248. Execution. — It is intended that this plate should be tinted, 

though, on account of the difficulty of procuring adequate engraved 
fac-similes of tinted hand-made drawings, it is here shown only as 
a finished line drawing, and as such, explains itself, after observing 
that as the left hand abutment is shown in elevation, it is dotted 
below the ground ; while, as the right hand abutment is shown in 
section, it is made wholly in full lines, and earth is shown only at 
each side of it. 

The usual conventional rule is, to fill the sectional elevation of a 
stone wall with wavy lines; but where other marks serve to distin- 
guish elevations from sections, as in the case just described, this 
labor is unnecessary. 

The following would be the general order of operations, in case 
this drawing were shaded. 

a. Pencil all parts in fine faint lines. 

b. Ink all parts in fine lines. 

c. Grain the wood work with a very fine pen and light indian ink, 



110 WOODEN STEUCTUEES. 

the sides of timbers as seen on a newly-planed board, the ends of 
large timbers in rings and radial cracks, and the ends of planks in 
diagonal straight lines. See also the figures at y, where the lines 
of graining outside of the knots, are to extend throughout the 
tie beam. 

d. Tint the wood work — the sides with pale clear burnt sienna, 
the ends with a darker tint of burnt sienna and indian ink. 

e. Tint the abutments, and other stone work, with prussian blue 
mixed with a little carmine and indian ink, put on in a very light 
tint. 

f. Grain the abutments in waving rows of fine, pale, vertical lines 
of uniform thickness, about one sixteenth of an inch long, leaving 
the upper and left hand edges of the stones blank, to represent the 
mortar. The part of the left hand abutment which is under ground 
is dotted only, as in the plate. 

g. Grain the canal walls and paving, as shown in the plate, to 
indicate boulder rubble. 

h. Shade the piles roughly, they being roughly cylindrical; tint 
them with' pale burnt sienna, and the shoe, S, with prussian blue, 
the conventional tint for iron. 

i. Rule the water in blue lines, distributed as in the figure. 

j. Tint the dirt in fine horizontal strokes of any dingy mixture, in 
which burnt sienna prevails, in the parts above the water, and ink, 
in the muddy parts below the water, and then add, or not, the pen 
strokes shown in the plate, to represent sand, gravel, &c. 

k. Place heavy lines on the right hand and lower edges of all 
surfaces, except Avhere such lines form dividing lines between two 
surfaces in the same plane. A heavy line on the under side of the 
floor planks, indicates that those planks project beyond the tie 
beam A. 

249. Ex. 5°. Construction from a Model. PI. XIV., Figs. 
127, 128. General Description. — This plate contains two elevations 
of an architectural Model. It is introduced as affording excellent 
practice in tinting and shading large surfaces, and useful elementary 
studies of shadows. The construction of these elevations from 
given measurements is so simple, that only the base and several 
centre lines need be pointed out. 

QR is the ground line. ST is a centre line for the flat topped 
tower in Fig. 127. TJV is a centre line for the whole of Fig. 128, 
except the left hand tower and its pedestal. WX is a centre line 
for the tower through which it passes. YZ is the centre line for 



WOODEN STKUCTUKES. Ill 

the roofed tower in Fig. 127.* The measurements are recorded in 
full, referred to the centre lines, base line, and bases of the towers, 
which are the parts to be first drawn. 

250. Graphical construction of the shadows. 

1°. The roof, D — D'D", casts a shadow on its tower. The point, 
EE', casts a shadow where the ray, Ee, pierces the side of the tower, 
e is one projection of this point ; e\ the other projection of the same 
point, is at the intersection of the line ee"e' with the other projection, 
EV, of the ray. The shadow of a line on a parallel plane (162) is 
parallel to itself, hence e'f\ parallel to E'F', is the shadow of E'F'. 

The shadow of DE— D'E' joins e' with the shadow of D— D'. 
The point d, determined by the ray Dc?, is one projection of the 
latter shadow ; the other projection, d\ is at the intersection of 
dd"d' with the other projection, T)'d\ of the ray. d' is on the side 
of the tower, produced, hence e'd' is only a real shadow line from 
e' till it intersects the edge of the tower. 

Remembering that the direction of the light is supposed to 
change with each position of the observer, so that as he faces each 
side of the model, in succession, the light comes from left to right 
and from behind his left shoulder, it appears that the point, DD", 
casts a shadow on the face of the tower, seen in Fig. 128, and that 
D"d"" will be the position of the ray, through this point, on Fig. 
127. The point d'" is therefore one projection of the shadow of 
DD". The other is at d'"\ the intersection of the lines d"d'" with 
Dd"'\ the other projection of the ray. Likewise EE" casts a 
shadow, e"'e"", on the same face of the tower, produced. DD'", 
being parallel to the face of the tower now being considered, its 
shadow, d""q, is parallel to it. The line from d"" towards e'" ', 
till the edge of the tower, is the real portion of the shadow of 
DE— D"E". 

251. From the foregoing it will be seen how most of these sha- 
dows are found, so that each step in the process of finding similar 
shadows will not be repeated. 

2°. The body of the building — or model — casts a shadow on the 
roofed tower, beginning at AA' (160). The shadow of BB' on the 
side of this tower is bb\ found as in previous cases, and A'b' is the 
shadow of AB — A'B'. From b' downwards, a vertical line is the 
shadow of the vertical corner edge of the body of the model upon 
the parallel face of the tower. 

3 8 . The line CO" — C, which is perpendicular to the side of the 
roofed tower, casts a shadow, CV, in the direction of the projection 
of a ray of light on the side of the tower. 



112 WOODEN STRUCTURES. 

4°. In Fig. 127, a similar shadow, s't', is cast by the edge s' — ss" 
of the smaller pedestal. 

5°. In Fig. 128, is visible the curved shadow, c"rg, cast by the 
vertical edge, at c'\ of the tower, on the curved part of the pedestal 
of the tower. The point g is found by drawing a ray, G'C — Qg, 
which meets the upper edge of the pedestal at gQ'. The point c"-> 
the intersection of the edge of the tower with the curved part of 
the pedestal, is another point. Any intermediate point, as r, is 
found by drawing the ray RV; r' is then one projection of the 
shadow of R'R, and the other is at the intersection of the line r'r 
with the other projection Rr of the ray. These are all the shadows 
which are very near to the objects casting them. 

6°. The flat topped tower casts a shadow on the roof of the body. 
The upper back corner, HH', casts a shadow on the roof,- of which 
h is one projection and h' the other. The back upper edge H — HT 
being parallel to the roof, the short shadow h'h", leaving the roof at 
h", is parallel to HT. The left hand back edge HJ — H'J' casts a 
shadow on the roof, of which hh! is one point. The point JJ' casts 
a shadow^)*' on the roof produced, h'f is therefore a real shadow 
only till it leaves the actual roof at u. 

7°. The same tower casts a shadow on the vertical side of the 
body, of which j"j"\ found as in previous cases, and u, are points. 
The upper back point, KK', of the shaft of the tower, casts a sha- 
dow, kk\ which is joined with J'", giving the shadow of JK — J'K'. 
From Jc\ h'l is the vertical shadow line of the left hand back edge 
of the flat topped tower on the parallel plane of the side of the body 
of the model. 

8°. The same tower casts a shadow on the curved — cylindrical — 
part of the pedestal. To find the point m', of shadow, draw a ray, 
MC, Fig. 128, intersecting the upper edge of the pedestal at C, 
which is therefore one projection of the shadow of the point MM'. 
The other projection, m', of the same shadow, is at the intersection 
of the other projection, Wm\ of the ray, with the other projection, 
m'C\ of the edge of the pedestal. The point of shadow, nn\ cast 
by the point W, is similarly found, and so is the point oo\ cast by 
the point 00' of the front right hand edge of the tower. Make 
ra'm" — n'o\ and find intermediate points, v'v", as rr' was found, 
and the curved shadow on the cylindrical part of the pedestal will 
then be found. 

9°. From n! and o\ vertical lines are the shadows of opposite 
diagonal edges of the tower, on the vertical face of the main 
pedestal. 



PL. XII. 




s 




[XI 1X1 IXIXI 1X1 IXl IXI IXI IXI M 1X1 





WOODEN - STRUCTURES. 113 

10°. This flat topped tower also casts a shadow on the side of the 
roofed tower. The right back corner, H — I', of the top, casts the 
shadow h'"h"" on the side of the roofed tower, through which the 
shadow line, h""x is drawn, parallel to the line H — HT which casts 
it. The right hand top line, T — IH, being perpendicular to the 
plane of the sides of this tower, casts the shadow h""i' upon it, 
parallel to the projection of a ray of light. (162.) This shadow line 
is real, only till it leaves the tower at 2/ — i' being in the plane of 
the side of the tower produced — and it completes all the shadows 
visible in the two elevations. 



8 



CHAPTER ni. 

IRON CONSTRUCTIONS. 

252. Ex. 6°. A Railway Track. PI. XV., Figs. 129-134. 
Mechanical construction, c&c. — It may be thought an oversight 

to style this plate the drawing of a railroad track ; but taking the 
track alone, or separate from its various special supports, as bridges, 
&c, its graphical representation is mainly summed up in that of two 
parts ; first, the union of two rails at their joints; second, the inter- 
section of two rails at the crossing of tracks, or at turn-outs. The 
fixture shown in Fig. 129, placed at the intersection of two rails to 
allow the unobstructed passage of car wheels, in either direction on 
either rail, is called a " Frog." Let y and z be fragments of two 
rails of the same track, then the side H/" of the point of the frog, 
and the portion h h' of its side flange, B, are in a line with the 
edges, denoted by dots, of the rails y and z, so that as the wheel 
passes either Avay, its flange rolls through the groove, I, without 
obstruction. When the wheel passes from y towards z there is a 
possibility of the flange's being caught in the groove, J, by dodg- 
ing the point,/ 1 . To guard against this, a guard rail, g g, is placed 
near to the inside of the other rail, supposed to be on the side of 
the frog towards Fig. 132, as shown in the small sketch, Fig. 132, 
which prevents the pair of wheels, or the car-truck, from working 
so far towards the flange, B, as to allow the flange of the wheel to 
run into the groove, J, and so run off the track. ¥f, and the por- 
tion, 1 1', of the flange, A, are in a line with the inner edge of the 
rail of a turn-out, for instance, the opposite rail being on the side of 
the frog towards the upper border of the plate, as shown in Fig. 
132. Hence the flange of a car wheel in passing in either direc- 
tion on the turn-out, passes through the groove, J, and is prevented 
from running into the groove, I, by a guard rail, near the inner 
edge of the opposite turn-out rail, as at U, Fig. 132. 

253. Fig. 130 represents the under side of the right hand portion 
of the frog, and shows the nuts which secure one of the bolts which 
secure the steel plates, as D, E ; bolts whose heads, as at u and v, 
are smooth and sunk into the plates so that their upper surfaces are 



IRON CONSTRUCTIONS. 115 

flush. It will be seen that there are two nuts on each bolt, as at 
D', on the bolt u — DD', which appears below the elevation, since it 
occurs between two of the cross-ties (sleepers) of the track. The 
nuts, as L, belonging to the bolt, b", which are in the chairs, q'p\ 
w\ x\ are sunk in cylindrical recesses in the bottom of the frog, so 
as not to interfere with the cross-tie on which the surface, L, rests. 
The extra nut is called a check or "jam" nut. When screwed on 
snugly it wedges the first nut and itself also against the threads of 
the screw, so that the violent tremulous motion to which the frog 
is subjected during the rapid passage of heavy trains cannot start 
either of them. 

In the end elevation, Fig. 131, A is the recess in the chair x x\ 
fitted for the reception of the rail, and B is the end of a rail in its 
place, as shown at y in the plan. 

254. Graphical Construction. — From the above description it 
follows that the whole length of the frog depends on the shape of 
the part H^F, and the distance between this part and the side 
rails, as c I. In the present example a c ==.!•' — 11* and cf= 20*. e d 
is 11" and nh is 2" from ~Ff. Having these relations given, and 
knowing that the lines at the extreme ends are perpendicular to the 
rails at those ends, the several figures of the frog can be constructed 
from the given measurements, without further explanation. 

255. Fig. 133, is an isometrical drawing — scale y 1 ^ — of a recent 
somewhat elaborate and very secure mode of connexion of the com- 
mon solid or H rails. A section of the Boston and Worcester R.R. 
recently (Nov. 1857) laid in this manner, allows the cars to ride 
over it* now, 1861, with great smoothness of motion and freedom 
from the loud clack which accompanies the use of ordinary chairs. 
A, A, A, are the sleepers (cross-ties), D is a stout oak plank, perhaps 
six feet long, resting on three sleepers, and fitted to the curved 
side of the rail, as shown at d. This plank is on the outside of 
the track. On the inner side the rails are spiked in the usual way 
with hook-headed spikes s s s, of which those at the joint, r, pass 
through a flat wrought iron plate, P, which gives a better bearing 
to the end of the rail, and prevents dislocation of parts. Each 
plank, as D, is bolted to the rail by four horizontal half inch bolts, 
6, 5, b, 5, furnished with nuts and washers on the further side of D 
(not seen). 

Note. — A recent (1860) modification of the above construction 
consists in substituting for the plate P, a short piece or strap of 
iron fitted to the surface of the inside of the rail, and through 
which the two bolts bb, next to the joint, pass. 



116 IRON CONSTRUCTIONS. 

Sometimes, indeed, such a short strap is placed on each side 
of the rail, and then the beam D is dispensed with. But wood is 
deemed preferable, since its elasticity renders the bolts, b. less 
likely to work loose than in a rigid iron construction. 

In connection w 7 ith a very solidly constructed section of the above 
railroad, on the plan just described, the experiment was made 
of having the track break joints. That is, a joint, as a, Fig. 132, 
on one rail of a track, is placed opposite the centre of the rail be of 
the other line of the same track. 

As a track always tends to settle at the joints, a jumping motion 
is induced in a passing train, which perhaps may be thought to be 
less violent if only on one rail at a time. A reduced jumping on 
the rail would in turn diminish the tendency to settle at the joints. 

256. Graphical Construction. — Three lines through X, making 
angles of 60° with each other, will be the isometric axes. Remem- 
bering that it is the relative position of the lines which distinguishes 
an isometrical drawing, we can place XX' parallel to the louver 
border, and thus fill out the plate to better advantage. The rail 
being 4" wide at bottom, and 4" high, circumscribe it by a square 
~Kcan, from the sides of which, or from its vertical centre line, lay 
off, on isometric lines, the distances to the various points on the rail. 
Thus, let the widest part of the rail, near the top, be 3" across, 
and \ an inch below the top ac. Let the width at the top be 2", 
and at the narrowest part 1"; and let the mean thickness of the 
lower flange be f ". The sides of the rail are represented by the 
bottom lines-at XX', and the tangents each side of R, to the curves 
of the section. Let the plank D be 6" wide, and 4" high. All 
the lines of the spikes, ss, are isometrical lines except their top 
edges, as st. The curve at the joint r, and at X', are similar to the 
corresponding parts of the section at X. 

To secure ease of graphical construction, let the bolt heads, 5, 
<fcc, be placed so that their edges shall be isometric lines. 

25V. Fig. 134, is a plan and end elevation of a heavy cast-iron 
chair lately (Nov. 1857) introduced upon the Troy and Boston R.R. 
On the outside of the rail, the top, ab, of the chair is flush with 
the top, be, of the rail, and thus forms essentially a continuous rail, 
to do which, while retaining the simplicity of the solid rail, is a 
thing to be desired. 

258. Ex. 1°. The Hydraulic Ram. In order to give an iron 
construction, from the department of machinery, so as to render 
this volume a more fit elementary course for the machinist as well 



IEON COXSTPwUCTIONS. 117 

as for the civil engineer, a simple and generally useful structure, 
viz. a hydraulic ram, has been chosen, as a fit example with which 
to close the present volume. 

This machine is designed to employ the power of running water 
to elevate water to any desired height. 

PI. XVI., Figs. 135-137, shows a hydraulic ram, of highly 
approved construction, and of half the full size. 

259. Jlechanical construction. — FF — F'F' are feet to support 
the machine. These are screwed to a floor or other firm support. 
AB — A'B'B' is the inlet pipe, opening into the air chamber C, at a — a'b r 
and ending at del — d'd' — d"d" the opening in the top of the waste 
valve chamber, E — E' — E". At a — a'b' is the opening as just 
noticed from the inlet pipe into the air chamber C (not seen in the 
plan). This opening is controlled by a leather valve ee\ weighted 
with a bit of copper e''e"\ and is fastened by a screw ti'h'", and 
an oblong washer g'g. At N" and H are the extremities of two 
outlet pipes leading from the air chamber at F"F"'. Either one, 
but not both of these outlet pipes together, may be used, as one 
of the exchangeable flanges, H' is solid, while the other is per- 
forated, as seen at M', Fig. 137. The air chamber is secured by 
bolts passing through its flange f'f\ through the pasteboard or 
leather packing, pp — p\ and the flange D — D'D' at cc. This 
flange, and part of the inlet pipe are shown as broken in the 
elevation, so as to expose the valve ee\ and the adjacent parts. 
LL' is a flange through which the inlet pipe passes, and this pipe 
is slit and bent over the inuer edge of the aperture in LL', forming 
a flange, which presses against a leather packing, tt\ and makes a 
tight joint. The outlet pipes are secured in the same way. At 
uu — u' are the square heads of bolts which fasten the flanges to 
the projections UU — U'. K — K' is a shelf bearing the waste 
valve chamber, E — E'E", and the adjacent parts. W — W is the 
flange of this valve chamber, secured by two bolts at vv" — v\ 
which pass through the leather packing y. h'h" is the waste 
valve, perforated with holes, cc, to allow water to flow through it. 
mm' is the valve stem, d'd'k'h' is a perforated standard serving 
as a guide to the valve stem, and also as a support to the hollow 
screw s. n is a rest, secured to the valve stem by a pin p" . q" 
is a nut, part of which, qq\ is made hexagonal, r is a "jam'' 
nut (253). 

In the plan of this portion of the machine, the innermost circle 
is the top of the valve stem; next is the body of the valve stem ; 
next, the top of the rest; next, the bottom of the same; next, the 



118 IRON CONSTRUCTIONS. 

nut q" ; and outside of that, and resting on the top of the waste 
valve chamber, are the standards, dd. 

960. Operation. — Principles involved. — In the case of what 
might be called passive constructions ; that is mere stationary sup- 
ports, like bridges, &c, a knowledge of the construction of the 
parts enables one to proceed intelligently in making a drawing ; 
but, in the case of what may, in opposition to the foregoing, be 
called active constructions, or machines, a knowledge of their mode 
of operation is usually essential to the most expeditious and accu- 
rate graphical construction of them, because a machine consists 
of a train of connected pieces, so that a given position of any piece 
implies a corresponding position for every other part. Having, 
then, in a drawing, assumed a definite position for some important 
part, the remaining parts must be located from a knowledge of the 
machine, though drawn by measurements of the dimensions of that 
part. Only fixed bearings, and centres of motion, can properly be 
located by measurement, in machine drawing. 

The principles involved in the operation of the hydraulic ram 
may be summed up under three heads, as follows: 

261. I. Work. a. When a certain weight is moved through a 
certain space, a certain amount of work is expended. 

b. Thus ; when a quantity of water descends through a certain 
space, a certain amount of work is developed. 

c, As the idea of work involves the idea both of weight moved, 
and space traversed, it follows that works may be equal, while the 
weights and spaces may be unequal. Thus the work developed by 
a certain quantity of water, while descending through a certain 
height, may be equal to that expended in raising a portion of that 
w T ater to a greater height. 

262. II. Equilibrium, a. Where forces are balanced, or mutu- 
ally neutralized, they are said to be in equilibrium. Now the usual 
fact is, that when such equilibrium is disturbed, it does not restore 
itself at once, but gradually, by a series of alternations about the 
state of equilibrium. Thus a stationary pendulum, being swung 
from its position of equilibrium, does not, at the first returning 
vibration, stop at the lowest point, but does so only after many 
vibrations. 

b. Theoretically, these vibrations, as in the case of the pendulum, 
would never stop, but in practice the resistance of the air, friction, 
&c, make a continual supply of a greater or less amount of force 
necessary to perpetuate the alternations about the position or 
state of equilibrium. 



IRON CONSTRUCTIONS. 119 

203. III. A physical fact taken account of in the hydraulic ram, 
is, that water in contact with compressed air will absorb a certain 
portion of such air. 

264. Passing now more particularly to a description of the ope- 
ration of the hydraulic ram: 1°. Water from some elevated pond 
or reservoir flows into the machine, through the inlet pipe AA' 
and continues through the machine, and flows out through the holes 
in the waste valve h!h'\ pressing meanwhile against the solid parts 
of the roof of this valve, whose hollow form — open at the bottom 
— is clearly shown in Fig. 136. 

2°. Presently the water acquires such a velocity as to press so 
strongly against the roof of the waste valve, that this valve is lifted 
against the under side of the roof of its chamber which it fits 
accurately. 

3°. The water thus instantly checked, expends its acquired force 
in rushing through the valve e — e'e"' and in compressing the air in 
the air chamber C. 

4°. The holes F" or F'" of the outlet pipe, leading to an unob- 
structed outlet, the compressed air immediately forces the water 
out through the outlet pipe until, after a number of repetitions of 
this chain of operations, the portion of the water thus expelled 
from the air chamber is raised to a considerable height. 

5°. In accordance with the second principle, the flow of water from 
the air chamber does not cease at the moment when the confined 
air is restored to its natural density, but continues, so that — taking 
account also of the absorption of the air by the water at the time 
of compression — for a moment the air of the air chamber is more rare 
than the external atmosphere. Hence to keep a constant supply 
of air to the air chamber, a fine hole called a snifting hole, is punc- 
tured, as with a needle, at ss\ i.e., just at the entrance of the inlet 
pipe into the machine. Through this hole air enters, with a snift- 
ing sound, when the flow of water recommences, so as to supply 
the air chamber with a constant quantity of air. "When the waste 
valve is at the bottom of the chamber EE', the nut and "jam" 
are together at the bottom of the screw s\ and the valve is at 
liberty to make a full stroke. By raising the valve to its highest 
point and turning the nut and "jam'' to some position as shown 
in the figure, the stroke of the valve can be shortened at pleasure, 
and, at its lowest point, will be as far from the bottom of the 
chamber as the "jam," q", is above its lowest position. 

266. In practice, it is found that the strokes of the waste valve 
shortly become regular ; their frequency depending in any given 



120 IRON CONSTRUCTIONS. 

case on the height of the supply reservoir, the height of the ejected 
column, the size of the machine, the length of the stroke of the 
valve, &c. 

267. The proportion of water discharged into the receiving 
reservoir will also depend on the above named circumstances, 
being more or less than one third of the quantity entering the 
machine at A A'. In a machine by M. Montgolfier of France, said 
to be the original inventor — water falling 7^ feet, raised -g T of itself 
to a height of 50 feet. 

268. Graphical Construction. — Scale ; half the full size. a. Hav- 
ing the extreme dimensions of the plan, in round numbers 9" and 
12", proceed to arrange the ground line, leaving room for the plan 
below it. 

b. Draw a centre line, "NG, for plan and elevation, about in the 
middle of the width of the plate. 

c. Draw a centre line, AK, for the plan, parallel to the ground line. 

d. Exactly 4£" from the centre line NO, draw the centre line vv" 
— K'm' for the waste valve chamber and parts adjacent. 

e. With the intersection, *, of the centre lines of the plan, as a 
centre, draw circles having radii of If" and 3^" respectively, and 
through the same centre, draw diagonals, as cc. 

f. On the centre line, NO, are the centres of the circles, F"F'", 
whose circumferences come within T T ¥ of an inch of the inner one 
of the two circles just drawn. 

g. Draw the valve, e, the copper weight e", the screw end, A, 
and the nut and oblong washer, h" and g. 

h. Locate, at once, the centres of all the small circles, cc, &c, by 
the intersection of arcs, J'' from the circle pp having * for a centre, 
w T ith the diagonals ; then proceed to draw these circles. 

i. Draw the projections, as U, drawing the opposite ones simul- 
taneously, and using an auxiliary end view of the nuts u, as often 
explained before. 

j. Draw the feet, F, with their grooves, F, and bevel edged screw 
holes, L". 

h. In drawing the shelf, K, and flange W, the intersection of the 
centre lines BK and m'm, is the centre for the curves which inter- 
sect the centre line AK; the-corners, 1 1, of the nuts, v,v'\ are the 
centres for the curves that cross the centre line, vv" ; and the 
remaining outlines of the shelf are tangents to the arcs thus drawn, 
and those of the flange are lines sketched in so as to give curves 
tangent to the arcs already drawn, and short straight lines parallel 



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IKON CONSTRUCTIONS. 121 

I. The remaining circles and larger hexagon, u\ of this portion 
of the plan, have the intersection of the centre lines for a centre ; and 
may be drawn by measurements independently of the elevation, or 
by projection from the elevation, after that shall have been 
finished. 

269. Passing to the elevation: — 

a. Construct, at one position of the T square, the horizontal lines 
of both feet ; then the horizontal lines of the nuts u\ and flange I/, 
and projection U' ; with the horizontal lines of the floor of the air 
chamber and adjacent parts. 

b. Project up from the plan the vertical edges of the feet, F'F', 
the flange, nut, and projection I/, u' and U', the valve e\ the cop- 
per e", the screw A", the washer #, the air chamber flange f'f\ and 
screw z. Break away the portion D — see plan — of the body of 
the machine, and the near wall of the water channel A'B'. Break 
away also the further wall of the water channel so as to show a 
section, H', of the further outlet pipe, H — see plan. Q is the centre 
of the spherical part of the air chamber to which the conical part 
is tangent. 

c. Draw all the horizontal lines of the waste valve chamber and 
parts adjacent. Make the edges of the threads of the screw straight 
and slightly inclined upwards toward the right. 

d. Project up from the plan, or lay off, by measurement, the 
widths of various parts through which the valve stem passes, and 
draw their vertical edges. 

Fig. 136 is a section of the waste valve chamber, showing part 
both of the interior and exterior of the waste valve. The dotted 
circles form an auxiliary plan of this valve, in which the holes have 
two radial sides, and two circular sides with x" as a centre. The 
top of the valve is conical, so that in the detail below, two of the 
sides of the hole ft, tend towards the vertex, x. At ft', one of these 
holes, of which there are supposed to be five, is shown in section. 

Fig. 137. The outlines of M, one of the outlet pipe flanges, are 
drawn by processes similar to those employed in drawing the shelf, 
K, in plan. 

270. Execution. As a line drawing, the plate explains itself. It 
would make a very beautiful shaded drawing and one that the 
careful student of the chapter on shading and shadows, would be 
able to execute with substantial accuracy, without further instruc- 
tion. 

THE END. 



pl.xiv: 




PL.XV. 




D1V.Y. 




pl. in. a. 




I'L.III A 




c^ 



